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Limits

Update: As of September 2022, we have much more interactive ways for you to learn about the foundational concept of Limits, making heavy use of Desmos graphing calculators. Please visit our Limits Chapter to really get this material down for yourself. If you’re wondering why these tactics work the way they do, start on this screen in that Chapter. It’s all free, and designed to help you do well in your course.

If you just need practice with limits problems for now, previous students have found what’s below super-helpful. And if you have questions, please ask on our Forum!

You probably already understand the basics of what a limit is, and how to find one by looking at the graph of a function. So we're going to jump right into where most students initially have some trouble: how to actually compute a limit in homework and exam problems. We'll illustrate the six tactics you must know, and then let you practice each.


1. Substitution
2. Factoring
3. Conjugates
4. Use Algebra
5. Trig Functions

If you get these down, you'll be able to easily solve almost any problem you're given.

(Problems where $x \to \infty$ deserve their own topic: Limits at Infinity.)

 
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Limits Tactic #1: Substitution
This is the first thing you should always try: just plug the value of x into f(x). If you obtain a number (and in particular, if you don't get  $\dfrac{0}{0}$), you have your answer and are finished. In that case, these problems are completely straightforward.

Example.
Find $\displaystyle{\lim_{x \to 2}(x^3-5x + 7)}$.

Solution. \begin{align*} \lim_{x \to 2}(x^3-5x + 7) &= (2)^3 -5(2) + 7 \\ &= 8 -10 + 7 = 5 \quad \cmark \end{align*}
Practice this (simple!) tactic in the next few problems. The solutions are immediately available using the Show/Hide Solution toggle. #5 illustrates an important point.
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Limits: Substitution Practice — #1
Find $\displaystyle{ \lim_{x \to -2}\left(x^3 +4\right) }$. \begin{array}{lllll} \text{(A) }-4 && \text{(B) }12 && \text{(C) }-12 && \text{(D) }-8 && \text{(E) None of these} \end{array}
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\begin{align*} \lim_{x \to -2}\left(x^3 +4\right) &= (-2)^3 +4 \\[4px] &= -8 +4 \\[4px] &= -4 \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Limits: Substitution Practice — #2
Find $\displaystyle{\lim_{x \to 4}\sqrt{9+\sqrt{4x}}}$. \begin{array}{lllll} \text{(A) }\sqrt{13} && \text{(B) }\sqrt{11} && \text{(C) }5 && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
\begin{align*} \lim_{x \to 4}\sqrt{9+\sqrt{4x}} &=\sqrt{9+\sqrt{4(4) }} \\[4px] &= \sqrt{9+\sqrt{16}} = \sqrt{9+4} \\[4px] &= \sqrt{13} \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Limits: Substitution Practice — #3
Find $\displaystyle{ \lim_{x \to -1} \sqrt[3]{\dfrac{x-7}{x+28}}}$.
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\begin{align*} \lim_{x \to -1} \sqrt[3]{\dfrac{x-7}{x+28}} &= \sqrt[3]{\dfrac{(-1)-7}{(-1)+28}} \\[8px] &= \sqrt[3]{\frac{-8}{27}} \\[8px] &= \frac{-2}{3} \quad \cmark \end{align*}
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Limits: Substitution Practice — #4
Find $\displaystyle{ \lim_{x \to 1} }\dfrac{\ln x}{x}$. \begin{array}{lllll} \text{(A) }1 && \text{(B) }e && \text{(C) }0 && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
\begin{align*} \lim_{x \to 1}\dfrac{\ln x}{x} &= \dfrac{\ln (1) }{1} \\[12px] &= \frac{0}{1} \\[12px] &= 0 \implies \quad\text{ (C)} \quad \cmark \end{align*}
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Limits: Substitution Practice — #5
Find $\displaystyle{\lim_{x \to 1} \dfrac{3}{x-1} }$.
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Answer: This limit does not exist. \[ \lim_{x \to 1} \dfrac{3}{x-1} = \dfrac{3}{1-1} = \frac{3}{0} \] Because this limit is in the form  $\dfrac{\text{a non-zero number}}{0}$,  specifically  $\dfrac{3}{0}$,   the function “blows up,” as the graph shows:

Graph of 3/(x-1) goes to neg infty as approach x=1 from the left and to infty as approach x=1 from the right

As $x \to 1^-$ from the left, the curve heads down toward $-\infty$. And as $x \to 1^+$ from the right, the curve heads up toward $\infty$. Hence this limit does not exist. $\quad \cmark$
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TIP
Tips icon As the preceding problem illustrates, when you substitute the limit-value into the function, if you obtain   \(\frac{\text{a non-zero number}}{0} \),   then you know immediately that the limit does not exist (DNE):

\[\text{If} \quad \lim_{x \to a}f(x) = \frac{\text{a non-zero number}}{0}\text{,}\] \[\text{then the limit does not exist (DNE).}\]
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Limits: Substitution Practice — #6
Find $\displaystyle{ \lim_{x \to \pi/2} } \cos x$. \begin{array}{lllll} \text{(A) }1 && \text{(B) }0 && \text{(C) }\dfrac{1}{\sqrt{2}} && \text{(D) DNE } && \text{(E) None of these} \end{array}
Show/Hide Solution
Recall that $\cos \left(\dfrac{\pi}{2} \right) = 0$. Hence \begin{align*} \lim_{x \to \pi/2}\cos x &= \cos \left(\dfrac{\pi}{2} \right) \\[8px] &= 0 \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Limits: Substitution Practice — #7
Find $\displaystyle{\lim_{x \to \pi /4}  } \sin x$.
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Answer: $\dfrac{1}{\sqrt{2}} $   or   $\dfrac{\sqrt{2}}{2}$.  The answers are equivalent and both are correct.

Recall that $\sin \left(\dfrac{\pi}{2}\right) = \dfrac{1}{\sqrt{2}}$. Hence \begin{align*} \lim_{x \to \pi /4}\sin x &= \sin \left(\dfrac{\pi}{2}\right) \\[8px] &= \dfrac{1}{\sqrt{2}} \quad \cmark \end{align*}
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Limits Tactic #2: Factoring
Frequently (on homework and exams, anyway) when you try Substitution, you obtain the fraction  $\dfrac{0}{0}$.  When this happens, if you can factor the numerator and/or the denominator, do so. A common term will cancel, removing the problematic  $0$  in the denominator. Guaranteed. Example. Find $\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$. Solution. We first try substitution: \begin{align*} \lim_{x \to 2}\frac{x^2-4}{x-2} &= \frac{2^2 -4}{2 - 2} = \frac{0}{0} \end{align*} Because this limit is in the form of  $\dfrac{0}{0}$,  it is "indeterminate"—we don't yet know what it is. So let's factor the numerator: \begin{align*} \lim_{x \to 2}\frac{x^2-4}{x-2} &= \frac{2^2 -4}{2 - 2} = \frac{0}{0} \end{align*} Because this limit is in the form of  $\dfrac{0}{0}$,  it is "indeterminate"—we don't yet know what it is. So let's factor the numerator: \begin{align*} \lim_{x \to 2}\frac{x^2-4}{x-2} &= \lim_{x \to 2}\frac{(x+2)(x-2)}{x-2} \\[8px] \text{Ah, now we can cancel the } &\text{problematic term:} &\phantom{= \lim_{x \to 2}\frac{(x+2)(x-2)}{x-2}}\\[8px] &= \lim_{x \to 2}\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}} \\[8px] &= \lim_{x \to 2} \,(x+2) \\[8px] \text{And now easy Substitution }& \text{to finish:} \\[8px] &= 2 + 2 = 4 \quad \cmark \end{align*}
Practice the tactic of factoring to find the limit in the next few problems. These are straightforward once you learn to recognize what to do. Note: Every Calculus exam on limits that we've ever seen has at least one problem that requires this tactic.
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Limits: Factoring Practice — #1
$\displaystyle{\lim_{x \to 3} \frac{x-3}{x^2-9}}$
Show/Hide Solution
Answer: $\dfrac{1}{6}$

We always first try Substitution: $$\lim_{x \to 3} \frac{x-3}{x^2-9} = \frac{3-3}{9-9} = \frac{0}{0}$$ Because this limit is in the form of  $\displaystyle{\frac{0}{0}}$,  it is “indeterminate”—we don’t know what it actually equals. So we have more work to do.

The first thing you should always try when hitting such an indeterminate limit is to factor the numerator or denominator, and see if common term cancels. In this case we can factor the denominator: \begin{align*} \lim_{x \to 3} \frac{x-3}{x^2-9} &= \lim_{x \to 3}\frac{(x-3)}{(x-3)(x+3)} \\[12px] &= \lim_{x \to 3}\frac{\cancel{(x-3)}}{\cancel{(x-3)}(x+3)} \\[12px] &= \lim_{x \to 3} \frac{1}{x+3} \\[12px] &= \frac{1}{3+3} = \frac{1}{6} \quad \cmark \end{align*} This function thus has the same graph as $y=\dfrac{1}{x+3}$, except that there is a hole at $x=3$ where our function is undefined. See the graph.

1 over x+3 with a hole
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Limits: Factoring Practice — #2
Find $\displaystyle{ \lim_{x \to 5} } \, \frac{x-5}{x^2-25}$. \begin{array}{lllll} \text{(A) }10 && \text{(B) }\dfrac{1}{10} && -\text{(C) }-\dfrac{1}{10} && \text{(D) 0} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to 5} \frac{x-5}{x^2-25} \overset{?}{=} \frac{5-5}{5^2 -25} = \frac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So let’s factor the denominator: \begin{align*} \lim_{x \to 5} \frac{x-5}{x^2-25} &= \lim_{x \to 5} \frac{x-5}{(x+5)(x-5)} \\[4px] &= \lim_{x \to 5} \frac{\cancel{x-5}}{(x+5)\cancel{(x-5)}} \\[4px] &= \lim_{x \to 5}\frac{1}{x+5} \\[4px] &= \frac{1}{5+5} = \frac{1}{10} \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Limits: Factoring Practice — #3
Find $\displaystyle{ \lim_{x \to -3} }\frac{x^2 -9}{x+3}$. \begin{array}{lllll} \text{(A) }6 && \text{(B) }-3 && \text{(C) }-6 && \text{(D) } 3 && \text{(E) nonexistent} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to -3}\frac{x^2 -9}{x+3} \overset{?}{=} \dfrac{(-3)^2 – 9}{(-3) +3} = \dfrac{0}{0} \] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So let’s factor the numerator: \begin{align*} \lim_{x \to -3}\frac{x^2 -9}{x+3} &= \lim_{x \to -3}\frac{(x+3) (x-3)}{(x+3)} \\[8px] &= \lim_{x \to -3} \frac{\cancel{(x+3)}(x-3)}{\cancel{(x+3)}} \\[8px] &= \lim_{x \to -3} (x-3) \\[8px] &= (-3) -3 = -6 \quad \text{(C)} \quad \cmark \end{align*}
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Limits: Factoring Practice — #4
Find $\displaystyle{ \lim_{x \to -4} }\dfrac{x^2-16}{x-4}$.
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Answer: 0

This is a little bit of a “trick question” since it’s listed as “Factoring Practice,” but in fact we don’t have to factor to find the answer. Instead, simple Substitution suffices: \[\lim_{x \to -4}\dfrac{x^2-16}{x-4} = \dfrac{(-4)^2 -16}{(-4) -4} = \dfrac{16 – 16}{-8} = \dfrac{0}{-8} =0 \quad \cmark\] That’s it; that’s the answer: 0.
The problem looks like it requires factoring, and factoring the numerator, canceling, and then substituting also leads to the correct answer of 0 . . . but we think it’s best to avoid taking extra steps when you don’t need to, since you risk making a simple algebra mistake that leads to a wrong answer when you do so. So remember: always try Substitution first!
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Limits: Factoring Practice — #5
Find $\displaystyle{ \lim_{x \to 1} }\frac{x^2 + 4x -5}{x^2 - 1}$. \begin{array}{lllll} \text{(A) }6 && \text{(B) }3 && \text{(C) }-3 && \text{(D) }-6 && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} \overset{?}{=} \frac{1 +4(1) – 5}{1-1} = \frac{0}{0}\] Because this limit is in the form of  $\dfrac{0}{0}$,  it is indeterminate—we don’t yet know what it is. So let’s factor the numerator and the denominator:

\begin{align*} \lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} &= \lim_{x \to 1} \frac{(x+5)(x-1)}{(x-1)(x+1)}\\[8px] &= \lim_{x \to 1} \frac{(x+5)\cancel{(x-1)}}{\cancel{(x-1)}(x+1)} \\[8px] &= \lim_{x \to 1}\frac{x+5}{x+1} \\[8px] &= \frac{1+5}{1+1} = \frac{6}{2} \\[8px] &=3 \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Limits: Factoring Practice — #6
Find $\displaystyle{ \lim_{x \to 0} }\,\frac{3x^3 -4x}{7x^3 +5x}$. \begin{array}{lllll} \text{(A) }-\dfrac{4}{5} && \text{(B) }-\dfrac{1}{12} && \text{(C) }\dfrac{3}{7} && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to 0} \frac{3x^3 -4x}{7x^3 +5x} \overset{?}{=} \frac{0 – 0}{0+0} = \frac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We notice that every term has an x in it, so let’s factor that out: \begin{align*} \lim_{x \to 0} \,\frac{3x^3 -4x}{7x^3 +5x} &= \lim_{x \to 0}\, \frac{x(3x^2 -4)}{x(7x^2+5)} \\[8px] &= \lim_{x \to 0} \frac{\cancel{x}(3x^2 -4)}{\cancel{x}(7x^2+5)} \\[8px] &= \lim_{x \to 0} \frac{(3x^2 -4)}{(7x^2+5)} \\[8px] &=  \frac{(0 -4)}{(0+5)} \\[8px] &= -\frac{4}{5} \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Limits: Factoring Practice — #7
Find $\displaystyle{ \lim_{x \to 1/3} }\,\dfrac{3x - 1}{1 - 9x^2}$. \begin{array}{lllll} \text{(A) }-\dfrac{1}{3} && \text{(B) }\dfrac{1}{2} && \text{(C) }-\dfrac{1}{2} && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to 1/3} \,\dfrac{3x – 1}{1 – 9x^2} \overset{?}{=} \dfrac{3\left(\dfrac{1}{3} \right) -1  }{1 – 9 \left(\dfrac{1}{3} \right)^2} = \dfrac{1-1}{1-1} = \dfrac{0}{0}\] Because this limit is in the form $\dfrac{0}{0}$, it is indeterminiate—we don’t yet know what it is.  So let’s factor the denominator: \begin{align*} \lim_{x \to 1/3}\, \dfrac{3x – 1}{1 – 9x^2} &= \lim_{x \to 1/3} \dfrac{3x – 1}{(1 -3x)(1+3x)} \\[8px] &= \lim_{x \to 1/3}\, \dfrac{-(1-3x)}{(1 -3x)(1+3x)} \\[8px] &= \lim_{x \to 1/3} \dfrac{-\cancel{(1-3x)}}{\cancel{(1 -3x)}(1+3x)} \\[8px] &= \lim_{x \to 1/3}\dfrac{-1}{1+3x} \\[8px] &= \frac{-1}{1+3\left(\dfrac{1}{3} \right) } = \frac{-1}{1+1} \\[8px] &=-\dfrac{1}{2} \implies \quad\text{ (C)} \quad \cmark \end{align*}
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Limits: Factoring Practice — #8
Find $\displaystyle{ \lim_{x \to 3} }\frac{x^2 -2x +5}{x-3}$.
Show/Hide Solution
Answer: This limit does not exist (DNE).

This is a little bit of a “trick question” since it’s under “Factoring Practice,” but we want to emphasize that you should always try Substitution first before you factor. In this case, Substitution shows that \[\lim_{x \to 3} \frac{x^2 -2x +5}{x-3} = \frac{3^2 -2(3) + 5}{3-3} = \frac{9-6+5}{0} = \frac{8}{0}\] Since this limit is in the form of  $\dfrac{\text{a nonzero number}}{0}$, the limit is nonexistent.  That is, the function “blows up” at $x = 3$.

Hence the limit does not exist (DNE) $\quad \cmark$ The function "blows up" in the limit as x goes to 3
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Limits: Factoring Practice — #9
Find $\displaystyle{\lim_{x \to 2} }\dfrac{x-2}{x^3 - 8}$.

Hint: Factor a cubic
Notice $x^3 – 8 = x^3 – 2^3.$ Then recall \[a^3 – b^3 = (a-b)\left(a^2 + ab + b^2 \right)\] You can multiply out the right hand side of the equation to verify the equality.
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\begin{array}{lllll} \text{(A) }\dfrac{1}{16} && \text{(B) }\dfrac{1}{8} && \text{(C) }\dfrac{1}{4} && \text{(D) }\dfrac{1}{12} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to 2} \dfrac{x-2}{x^3 – 8} \overset{?}{=} \dfrac{2-2}{2^3-8} = \dfrac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So we factor the denominator: \begin{align*} \lim_{x \to 2} \dfrac{x-2}{x^3 – 8} &= \lim_{x \to 2} \dfrac{x-2}{(x-2)(x^2 + 2x + 4)} \\[8px] &= \lim_{x \to 2} \dfrac{\cancel{x-2}}{\cancel{(x-2)}(x^2 + 2x + 4)} \\[8px] &= \lim_{x \to 2} \dfrac{1}{(x^2 + 2x + 4)}\\[8px] &= \frac{1}{2^2 + 2(2)+4} = \frac{1}{4 + 4 + 4} \\[8px] &= \frac{1}{12} \implies \quad\text{ (D)} \quad \cmark \end{align*}
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Limits: Factoring Practice — #10
[Challenging] For what value of k does $\displaystyle{ \lim_{x \to 3} }\frac{x^2 -2x +k}{x-3}$ exist?

Open for hint.
If Substitution yields the result $\dfrac{\text{a nonzero number}}{0},$ then we know immediately that the limit does not exist. Instead Substitution must yield the indeterminate result $\dfrac{0}{0}.$ Use this fact to determine k.
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Show/Hide Solution
Answer: $k = -3$

We see immediately that Substition $x=3$ will yield 0 in the denominator. And we know that if Substitition yields $\dfrac{\text{a nonzero number}}{0},$ then the limit does not exist, contrary to the requirements here.

Hence the numerator must equal zero when we substitute $x=3,$ such that Substition yields the indeterminate result $\dfrac{0}{0}.$ We thus know that \begin{align*} \lim_{x \to 3}[x^2 – 2x +k] &= 0 \\[8px] 3^2 -2(3) + k &= 0 \\[8px] 9-6 +k &= 0 \\[8px] 3 + k &= 0 \\[8px] k &= -3 \quad \cmark \end{align*}
Let’s double-check our answer by treating this as a new problem that asks us to find the limit with $k=-3$. That is, imagine we have a new problem that asks us to find $\displaystyle{ \lim_{x \to 3}\frac{x^2 -2x -3}{x-3} }$. As usual to proceed we factor the numerator: \begin{align*} \lim_{x \to 3}\frac{x^2 -2x -3}{x-3} &= \lim_{x \to 3} \frac{(x-3)(x+1)}{x-3}\\[8px] &= \lim_{x \to 3} \frac{\cancel{(x-3)}(x+1)}{\cancel{x-3}}\\[8px] &= \lim_{x \to 3} \, (x+1) = 4\\[8px] \end{align*}

The exact limit there doesn’t matter; the key point is that with $k=-3$, the limit exists because the $x-3$ in the denominator canceled.  And that’s what the problem demanded.  $\checkmark$

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Limits Tactic #3: Conjugates
If you try Substitution and obtain the fraction  $\dfrac{0}{0}$  and the expression has a square root in it, then rationalize the expression just like you practiced in Algebra. That is, multiply the numerator and the denominator by the conjugate of the part that has the square root. The following example illustrates.

Example. Find $\displaystyle{ \lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x}}$.

Solution. As always, we first try Substitution:

$$\lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x} = \frac{\sqrt{5} - \sqrt{5}}{0} = \dfrac{0}{0}$$ Since this limit is in the form of  $\dfrac{0}{0}$,  it is indeterminate—we don't yet know what it is. We have more work to do.

So let's try to rationalize the expression. The square-root part is $\sqrt{x+5} - \sqrt{5}$, so we multiply both the numerator and the denominator by the conjugate $\dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}}=1$:

\begin{align*} \lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x} &= \lim_{x \to 0}\dfrac{\sqrt{x+5} - \sqrt{5}}{x} \cdot \dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}} \\[8px] &= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \sqrt{x+5}\sqrt{5} - \sqrt{5}\sqrt{x+5} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \cancel{\sqrt{x+5}\sqrt{5}} - \cancel{\sqrt{5}\sqrt{x+5}} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{(x+5) - 5}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{x}{x[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{\cancel{x}}{\cancel{x}[\sqrt{x+5} + \sqrt{5}]} \\[8px] &= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\[8px] &= \dfrac{1}{\sqrt{0+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} \quad \cmark \end{align*} Tips icon Notice that when we multiplied by the conjugate, we multiplied out all of the terms in the numerator, because that's how we get rid of the square root. But we didn't multiply the terms in the denominator; instead we kept writing it as $x \left(\sqrt{x+5} + \sqrt{5} \right)$. That's because a few steps later the x canceled. Something similar will always happen, so in that early step don't multiply out the part that you didn't set out to rationalize. Instead just carry those terms along for a while, until you can cancel something.
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Limits: Conjugates Practice — #1
Find $\displaystyle{\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}}$.
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Answer: 6

We first try substitution: $$ \lim_{x \to 9} \frac{x-9}{\sqrt{x}-3} = \frac{9-9}{3-3} = \frac{0}{0} $$ Since this limit is in the form  $\dfrac{0}{0}$,  it is indeterminate and tells us only that we have more work to do.

So let’s rationalize the denominator by using the usual approach of multiplying by its conjugate $ \dfrac{\sqrt{x} + 3}{\sqrt{x} + 3} = 1$ :

\begin{align*} \lim_{x \to 9} \frac{x-9}{\sqrt{x}-3} &= \lim_{x \to 9} \left( \frac{x-9}{\sqrt{x}-3} \right) \left( \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \right) \\[12px] &= \lim_{x \to 9} \frac{(x-9)(\sqrt{x} + 3)}{(x – 9)} \\[12px] &= \lim_{x \to 9} \frac{\cancel{(x-9)}(\sqrt{x} + 3)}{\cancel{(x – 9)}} \\[12px] &= \lim_{x \to 9}\sqrt{x} + 3 \\[12px] &= \sqrt{9} + 3 = 6 \quad \cmark \end{align*}
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Limits: Conjugates Practice — #2
Find $\displaystyle{\lim_{x \to 1} \frac{\sqrt{x}-1}{x-1} }$.
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Answer: $\dfrac{1}{2}$

We first try Substitution: \[ \lim_{x \to 1} \frac{\sqrt{x}-1}{x-1} = \frac{\sqrt{1}-1}{1-1} = \frac{0}{0} \] Since this limit is in the form of  $\dfrac{0}{0}$  it is indeterminate—we don’t yet know what it is. So let’s try rationalizing the expression:

\begin{align*} \lim_{x \to 1} \frac{\sqrt{x}-1}{x-1} &= \lim_{x \to 1} \frac{\sqrt{x}-1}{x-1} \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} \\[8px] &= \lim_{x \to 1}\frac{x-1}{(x-1) \left( \sqrt{x}+1\right)} \\[8px] &= \lim_{x \to 1}\frac{\cancel{x-1}}{\cancel{(x-1)} \left( \sqrt{x}+1\right)} \\[8px] &= \lim_{x \to 1} \frac{1}{\left( \sqrt{x}+1\right)} \\[8px] &= \frac{1}{\left( \sqrt{1}+1\right)} \\[8px] &= \frac{1}{2} \quad \cmark \end{align*}
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Limits: Conjugates Practice —#3
Find $\displaystyle{ \lim_{h \to 0} \frac{\sqrt{25+h}-5}{h}}$. \begin{array}{lllll} \text{(A) }\dfrac{1}{5} && \text{(B) }\dfrac{1}{10} && \text{(C) }5 && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try Substitution: \[ \lim_{h \to 0} \frac{\sqrt{25+h}-5}{h} \overset{?}{=} \frac{\sqrt{25} – 5}{0} = \frac{0}{0} \] Since this limit is in the form of  $\dfrac{0}{0}$,  it is indeterminate—we don’t yet know what it is. So let’s use our conjugate tactic: \begin{align*} \lim_{h \to 0} \frac{\sqrt{25+h}-5}{h} &= \lim_{h \to 0} \frac{\sqrt{25+h}-5}{h} \cdot \frac{\sqrt{25+h}+5}{\sqrt{25+h}+5} \\[8px] &= \lim_{h \to 0} \frac{\sqrt{25+h}\sqrt{25+h} + \sqrt{25+h} (5) -5\sqrt{25+h} -(5)(5) }{h\left(\sqrt{25+h}+5 \right)} \\[8px] &= \lim_{h \to 0} \frac{(25 +h)-25}{h\left(\sqrt{25+h}+5 \right)} \\[8px] &= \lim_{h \to 0} \frac{h}{h\left(\sqrt{25+h}+5 \right)} \\[8px] &= \lim_{h \to 0} \frac{\cancel{h}}{\cancel{h}\left(\sqrt{25+h}+5 \right)} \\[8px] &= \lim_{h \to 0} \frac{1}{\sqrt{25+h}+5 } \\[8px] &= \frac{1}{\sqrt{25} + 5} \\[8px] &= \frac{1}{5+5} = \frac{1}{10} \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Limits: Conjugates Practice —#4
Find $\displaystyle{ \lim_{x \to 0} \frac{x}{\sqrt{x+7} - \sqrt{7}} }$. \begin{array}{lllll} \text{(A) }2\sqrt{7} && \text{(B) }-2\sqrt{7} && \text{(C) }\dfrac{1}{\sqrt{7}} && \text{(D) }\dfrac{1}{2\sqrt{7}} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: \[\lim_{x \to 0} \frac{x}{\sqrt{x+7} – \sqrt{7}} \overset{?}{=} \frac{0}{\sqrt{7} – \sqrt{7}}=\frac{0}{0}\] Since this limit is in the form $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is.  So let’s multiply the numerator and denominator by the conjugate of the square-root term divided by itself, $ \dfrac{\sqrt{x+7} + \sqrt{7}}{\sqrt{x+7} + \sqrt{7}} =1:$ \begin{align*} \lim_{x \to 0} \frac{x}{\sqrt{x+7} – \sqrt{7}} &= \lim_{x \to 0} \frac{x}{\sqrt{x+7} – \sqrt{7}}\cdot \frac{\sqrt{x+7} + \sqrt{7}}{\sqrt{x+7} + \sqrt{7}} \\[8px] &= \lim_{x \to 0}\frac{x \left( \sqrt{x+7} + \sqrt{7}\right)}{\sqrt{x+7}\sqrt{x+7} + \sqrt{x+7} \sqrt{7} – \sqrt{7}\sqrt{x+7} -\sqrt{7}\sqrt{7}} \\[8px] &= \lim_{x \to 0}\frac{x \left( \sqrt{x+7} + \sqrt{7}\right)}{(x+7) – 7} \\[8px] &= \lim_{x \to 0}\frac{x \left( \sqrt{x+7} + \sqrt{7}\right)}{x} \\[8px] &= \lim_{x \to 0}\frac{\cancel{x} \left( \sqrt{x+7} + \sqrt{7}\right)}{\cancel{x}} \\[8px] &= \lim_{x \to 0}\left( \sqrt{x+7} + \sqrt{7}\right) \\[8px] &= \sqrt{7} + \sqrt{7}\\[8px] &= 2 \sqrt{7} \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Limits: Conjugates Practice —#5
Find $\displaystyle{ \lim_{x \to 0} \left(\frac{1}{x \sqrt{1+x}} - \frac{1}{x} \right) }$.

Open for hint.
First put both terms over a common denominator. Then proceed as usual.
[collapse]
\begin{array}{lllll} \text{(A) }1 && \text{(B) }\dfrac{1}{2} && \text{(C) }-\dfrac{1}{2} && \text{(D) DNE} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try Substitution: $$\lim_{x \to 0} \left(\frac{1}{x \sqrt{1+x}} – \frac{1}{x} \right) \overset{?}{=} \frac{1}{0} – \frac{1}{0} \quad \text{??} $$ This is another form of an indeterminite limit: we don’t know what $\dfrac{1}{0} – \dfrac{1}{0}$ equals, so we have more work to do. Let’s try our usual approach, and first put the two terms over a common denominator, and then use the conjugate: \begin{align*} \lim_{x \to 0} \left(\frac{1}{x \sqrt{1+x}} – \frac{1}{x} \right) &= \lim_{x \to 0} \frac{1 -\sqrt{1+x} }{x \sqrt{1+x}} \\[8px] &= \lim_{x \to 0} \frac{1 -\sqrt{1+x} }{x \sqrt{1+x}} \cdot \frac{1 + \sqrt{1+x}}{1 + \sqrt{1+x}} \\[8px] &= \lim_{x \to 0} \frac{1 + \sqrt{1+x}\; – \sqrt{1+x}\; – \sqrt{1+x} \sqrt{1+x} }{x \sqrt{1+x}\left(1 + \sqrt{1+x} \right)} \\[8px] &= \lim_{x \to 0} \frac{1 -(1+x)}{x \sqrt{1+x}\left(1 + \sqrt{1+x} \right)} \\[8px] &= \lim_{x \to 0} \frac{-x}{x \sqrt{1+x}\left(1 + \sqrt{1+x} \right)} \\[8px] &= \lim_{x \to 0} \frac{-\cancel{x}}{\cancel{x}\sqrt{1+x}\left(1 + \sqrt{1+x} \right)} \\[8px] &= \lim_{x \to 0} \frac{-1}{\sqrt{1+x}\left(1 + \sqrt{1+x} \right)} \\[8px] &= \frac{-1}{\sqrt{1+0}\left(1 + \sqrt{1+0} \right)} \\[8px] &= \frac{-1}{1\left(1 +1 \right)} \\[8px] &= \frac{-1}{2} \implies \quad\text{ (C)} \quad \cmark \end{align*}
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Limits: Conjugates Practice —#6
Find $\displaystyle{ \lim_{x \to 3} \frac{\sqrt{x^2 + 7}-4}{x^2-9}}$. \begin{array}{lllll} \text{(A) }\dfrac{1}{8} && \text{(B) }\dfrac{1}{4} && \text{(C) }-\dfrac{1}{4} && \text{(D) undefined} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try Substitution: \[\lim_{x \to 3} \frac{\sqrt{x^2 + 7}-4}{x^2-9} \overset{?}{=} \frac{\sqrt{9 + 7}-4}{9-9} = \frac{\sqrt{16}-4}{9-9} = \frac{0}{0}\] Because this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. So let’s again use the conjugate divided by itself: \begin{align*} \lim_{x \to 3} \frac{\sqrt{x^2 + 7}-4}{x^2-9} &= \lim_{x \to 3}\frac{\sqrt{x^2 + 7}-4}{x^2-9}\cdot \frac{\sqrt{x^2 + 7}+4}{\sqrt{x^2 + 7}+4} \\[8px] &+\lim_{x \to 3}\frac{\sqrt{x^2 + 7}\sqrt{x^2 + 7} + 4\sqrt{x^2 + 7} -\sqrt{x^2 + 7}(4) -(4)(4)}{\left(x^2 -9 \right)\left(\sqrt{x^2 + 7}+4 \right)} \\[8px] &= \lim_{x \to 3} \frac{\left(x^2 + 7 \right)-16}{\left(x^2 -9 \right)\left(\sqrt{x^2 + 7}+4 \right)}\\[8px] &= \lim_{x \to 3} \frac{x^2- 9}{\left(x^2 -9 \right)\left(\sqrt{x^2 + 7}+4 \right)} \\[8px] &= \lim_{x \to 3} \frac{\cancel{x^2- 9}}{\cancel{\left(x^2 -9 \right)}\left(\sqrt{x^2 + 7}+4 \right)}\;\color{blue}{\text{[Amazing: cancellation always happens.]}} \\[8px] &= \lim_{x \to 3}\frac{1}{\sqrt{x^2 + 7}+4 } \\[8px] &= \frac{1}{\sqrt{9 + 7}+4 } \\[8px] &= \frac{1}{\sqrt{16}+4} = \frac{1}{4+4} = \frac{1}{8} \implies \quad\text{ (A)} \quad \cmark \end{align*}
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Limits: Conjugates Practice —#7
Find $\displaystyle{ \lim_{x \to 3}\frac{\sqrt{x+1} -2}{3-x}.}$ \begin{array}{lllll} \text{(A) }\dfrac{1}{6} && \text{(B) }-\dfrac{1}{6} && \text{(C) }\dfrac{1}{4} && \text{(D) nonexistent} && \text{(E) None of these } \end{array}
Show/Hide Solution
We first try substitution: \[ \lim_{x \to 3}\frac{\sqrt{x+1} -2}{3-x} \overset{?}{=} \frac{\sqrt{3+1} -2}{3-3} = \frac{\sqrt{4}-2}{3-3} = \frac{0}{0} \] Since this limit is in the form $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is.  So we again use the conjugate: \begin{align*} \lim_{x \to 3}\frac{\sqrt{x+1} -2}{3-x} &= \lim_{x \to 3}\frac{\sqrt{x+1} -2}{3-x} \cdot \frac{\sqrt{x+1} +2}{\sqrt{x+1} +2} \\[8px] &= \lim_{x \to 3} \frac{\sqrt{x+1}\sqrt{x+1}+ 2\sqrt{x+1} – 2 \sqrt{x+1} -4}{(3-x)(\sqrt{x+1}+2)} \\[8px] &= \lim_{x \to 3}\frac{(x+1)-4}{(3-x)(\sqrt{x+1}+2)}\\[8px] &= \lim_{x \to 3}\frac{x-3}{(3-x)(\sqrt{x+1}+2)} \\[8px] &= \lim_{x \to 3}\frac{-(3-x)}{(3-x)(\sqrt{x+1}+2)} \;\color{blue}{\text{[Factor out $-1$ to cancel terms]}}\\[8px] &= \lim_{x \to 3}\frac{-\cancel{(3-x)}}{\cancel{(3-x)}(\sqrt{x+1}+2)} \\[8px] &= \lim_{x \to 3}\frac{-1}{\sqrt{x+1}+2} \\[8px] &= \frac{-1}{\sqrt{3+1}+2} \\[8px] &= \frac{-1}{\sqrt{4}+2} = \frac{-1}{2+2} \\[8px] &= \frac{-1}{4} \implies \quad\text{ (E)} \quad \cmark \end{align*}
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Limits Tactic #4: Use Algebra (polynomial expansion, common denominator, ...)
For some problems, when you try Substitution and obtain  $\dfrac{0}{0}$,  you just use your algebraic skills to transform the expression into something where Substitution will work. For instance, if there's a quadratic, expand it. Or if you have some fractions in the numerator, put them over a common denominator.

The following problems illustrate.
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Limits: Use Algebra Practice — #1
Find $\displaystyle{\lim_{h \to 0}\dfrac{(h-5)^2 - 25}{h}}$. \begin{array}{lllll} \text{(A) }10 && \text{(B) }-10 && \text{(C) }5 && \text{(D) }-5 && \text{(E) None of these} \end{array}
Show/Hide Solution
As always, we first try Substitution: $$\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} \overset{?}{=} \dfrac{(0-5)^2 -25}{0} = \dfrac{25-25}{0} = \dfrac{0}{0}$$ Since the limit is in the form   $\dfrac{0}{0},$  it is indeterminate—we don’t yet know what is it. So we have to do some work to turn the expression into a different form that’s more helpful. Let’s try the simple move of expanding the quadratic in the numerator, and then see what happens: \begin{align*} \lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} &= \lim_{h \to 0}\dfrac{(h^2 -10h + 25) – 25}{h} \\[8px] &= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\[8px] &= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\[8px] &= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \quad \text{[Cool: cancellation.]}\\[8px] &= \lim_{h \to 0}(h – 10) \\[12px] &= 0\, – 10 \\[8px] &= -10 \implies \quad\text{ (B)} \quad \cmark \end{align*} Notice that the problematic h in the denominator cancelled out, just as we were hoping would happen. We can then use simple Substitution to finish.
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Limits: Use Algebra Practice #2
Find $\displaystyle{\lim_{x \to 4} \frac{x^{-1} - 4^{-1}}{x-4}}$.
Show/Hide Solution
Answer:  $-\dfrac{1}{16}$

It’s hard to see what’s going on with the negative exponents, so we’ll start by writing the terms in the numerator as fractions:

$\displaystyle{\lim_{x \to 4} \frac{x^{-1} – 4^{-1}}{x-4}}$\(=\)$\displaystyle{ \lim_{x \to 4} \frac{\frac{1}{x} – \frac{1}{4}}{x-4}}$
Now let’s put the terms in the numerator
over a common denominator, $4x$:
\(=\)$\displaystyle{\lim_{x \to 4} \frac{\frac{4-x}{4x}}{x-4}}$
\(=\)$\displaystyle{ \lim_{x \to 4} \left( \frac{4-x}{4x} \right) \left( \frac{1}{x-4} \right)}$
We can cancel some terms if we factor a
$-1$ out of the $4-x$:
\(=\)$\displaystyle{\lim_{x \to 4} \left( \frac{-(x-4)}{4x} \right) \left( \frac{1}{x-4} \right)}$
\(=\)$\displaystyle{\lim_{x \to 4} \left( \frac{-\cancel{(x-4})}{4x} \right) \left( \frac{1}{\cancel{(x-4)}} \right)}$
\(=\)$\displaystyle{\lim_{x \to 4} \frac{-1}{4x}}$
\(=\)$\displaystyle{\frac{-1}{4(4)} = \frac{-1}{16} \quad \cmark }$
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Limits: Use Algebra Practice #3
Find $\displaystyle{\lim_{h \to 0}\frac{(5+h)^{-1} -5^{-1}}{h}}$. \begin{array}{lllll} \text{(A) }\dfrac{1}{5} && \text{(B) }-\dfrac{1}{5} && \text{(C) }\dfrac{1}{25} && \text{(D) }-\dfrac{1}{25} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try Substitution: $$\lim_{h \to 0}\frac{(5+h)^{-1} -5^{-1}}{h} = \frac{(5+0)^{-1} – 5^{-1}}{0} = \frac{0}{0} $$ Since this limit is in the form of $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is. We have more work to do. So let’s put the terms in the numerator over a common denominator, and go from there: \begin{align*} \lim_{h \to 0}\frac{(5+h)^{-1} -5^{-1}}{h} &= \lim_{h \to 0}\frac{\dfrac{1}{5+h} -\dfrac{1}{5}}{h} \\[8px] &= \lim_{h \to 0}\frac{\dfrac{5}{5(5+h)} -\dfrac{(5+h)}{5(5+h)}}{h} \\[8px] &= \lim_{h \to 0}\frac{\dfrac{5 – (5+h)}{5(5+h)}}{h} \\[8px] &= \lim_{h \to 0}\frac{\dfrac{-h}{5(5+h)}}{h} \\[8px] &= \lim_{h \to 0}\frac{-h}{5(5+h)h} \\[8px] &= \lim_{h \to 0}\frac{-\cancel{h}}{5(5+h)\cancel{h}} \\[8px] &= \lim_{h \to 0}\frac{-1}{5(5+h)} \\[8px] &= \frac{-1}{5(5+0)} \\[8px] &= \frac{-1}{25} \implies \quad\text{ (D)} \quad \cmark \end{align*}
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Limits Tactic #5: Trig Functions
When your problem has a trig function in it (sin, cos, tan, ...), often you need to do some manipulation in order to find the limit. To help:

1. Remember your fundamental trig identities. For instance:

  • $\tan x = \dfrac{\sin x}{\cos x}$
  • $\csc x = \dfrac{1}{\sin x}$
  • $\sin^2(x) + \cos^2(x) = 1$
Our Handy Table of Trig Formulas and Identities could be useful.

2. Early in the semester, there are two "Special Limits" you just have to memorize:

\begin{align*} \text{I. } & \lim_{x \to 0}\frac{\sin(x)}{x} = 1 \\[16px] \text{II. } & \lim_{x \to 0}\frac{1-\cos(x)}{x} = 0 \end{align*}
The following problems illustrate how you'll frequently have to make use of these facts.
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Limits: Trig Practice #1
Find $\displaystyle{\lim_{x \to 0}\dfrac{1- \cos(x) }{\sin^2(x)} }$.

Open for hint.
Recall the trig identity \[\sin^2(x) + \cos^2(x) = 1\]
[collapse]
\begin{array}{lllll} \text{(A) }1 && \text{(B) }0 && \text{(C) } \dfrac{1}{2} && \text{(D) Does not exist} && \text{(E) None of these} \end{array}
Show/Hide Solution
As always, we first try Substitution: $$\lim_{x \to 0}\dfrac{1- \cos(x)}{\sin^2(x)} \overset{?}{=} \frac{1- \cos(0)}{\sin^2(0)} = \frac{1-1}{0} = \frac{0}{0}$$ Yet again we obtain  $\dfrac{0}{0}$,  that indeterminate result: we don’t yet know what the limit is. We have more work to do.

Seeing the $\sin^2(x)$ there in the denominator makes us think about the trig identity in the hint, $\sin^2(x) + \cos^2(x) = 1,$ which we can rewrite as $$ \sin^2(x) = 1 – \cos^2(x) $$ Hence \begin{align*} \lim_{x \to 0}\dfrac{1- \cos(x) }{\sin^2(x)} &= \lim_{x \to 0}\frac{1- \cos(x) }{1 – \cos^2(x)} \\[8px] &= \lim_{x \to 0}\frac{1- \cos(x) }{(1 – \cos(x))(1 + \cos(x))} \\[8px] &= \lim_{x \to 0}\frac{\cancel{1- \cos(x)} }{\cancel{(1 – \cos(x))}(1 + \cos(x))} \\[8px] &= \lim_{x \to 0} \frac{1}{1 + \cos(x)} \\[8px] &= \frac{1}{1+\cos(0)} \\[8px] &= \frac{1}{1+1} = \frac{1}{2} \implies \quad\text{ (C)} \quad \cmark \\[8px] \end{align*}
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Limits: Trig Practice #2
Find $\displaystyle{\lim_{x \to 0}\dfrac{\sin(5x)}{x} }$.
Show/Hide Solution
Answer: 5

The expression in the question reminds us of the first “Special Limit,” $$\lim_{x \to 0}\frac{\sin(x)}{x} = 1$$ But it isn’t quite the same, because in our expression the argument of sin that’s in the numerator (5x) doesn’t match what’s in the denominator (x). That is, since we have $\sin(5x)$ in the numerator, we need $5x$ in the denominator.

So let’s multiply the expression by  $\dfrac{5}{5}$, and then do some rearranging:

\begin{align*} \lim_{x \to 0}\dfrac{\sin(5x)}{x} &= \lim_{x \to 0} \frac{5}{5} \cdot \dfrac{\sin(5x)}{x} \\[8px] &= 5 \cdot \lim_{x \to 0} \frac{\sin(5x)}{5x} &\text{[Recall $\displaystyle{\lim_{x \to 0}\dfrac{\sin(\text{whatever})}{(\text{the same whatever})} =1 }$]}\\[8px] &= 5 \cdot 1 = 5 \quad \cmark \end{align*}
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Limits: Trig Practice #3
Find $\displaystyle{\lim_{x \to 0} \frac{\sin 3x}{\sin 5x} }$. \begin{array}{lllll} \text{(A) }\dfrac{5}{3} && \text{(B) }\dfrac{3}{5} && \text{(C) }\dfrac{1}{15} && \text{(D) Does not exist} && \text{(E) None of these} \end{array}
Show/Hide Solution
We first try substitution: $$\frac{\sin 3x}{\sin 5x} = \frac{\sin 0}{\sin 0} = \frac{0}{0} $$ Since this limit is in the form of $\dfrac{0}{0}$, it is indeterminiate — we don’t yet know what it is. We have more work to do. This limit reminds us somewhat of our special limit $$ \lim_{x \to 0} \frac{\sin\text{(whatever)}}{\text{(the same whatever)}} = 1$$ but clearly it isn’t currently in a form where we can use that special limit. So let’s do some manipulation to make it into that form: \begin{align*} \lim_{x \to 0}\frac{\sin 3x}{\sin 5x} &= \lim_{x \to 0}\left( \sin 3x \cdot \frac{3x}{3x} \right) \left(\frac{1}{\sin 5x} \cdot \frac{5x}{5x} \right) \\[8px] &= \lim_{x \to 0}\left(\frac{\sin 3x}{3x} \right) \left(\frac{5x}{\sin 5x} \right) \cdot \frac{3x}{5x} \\[8px] &= \lim_{x \to 0}\left(\frac{\sin 3x}{3x} \right) \left(\frac{5x}{\sin 5x} \right) \cdot \frac{3\cancel{x}}{5\cancel{x}} \\[8px] &= \left(\lim_{x \to 0}\frac{\sin 3x}{3x} \right) \left(\lim_{x \to 0}\frac{5x}{\sin 5x} \right) \cdot \frac{3}{5} \\[8px] &=(1)(1)\cdot \frac{3}{5} \\[8px] &= \frac{3}{5} \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Limits: Trig Practice #4
Find $\displaystyle{\lim_{x \to 0} \dfrac{1 - \cos (x)}{x \cos (x)} }$.
Show/Hide Solution
Answer: $0$

We first try Substitution:

$$ \lim_{x \to 0} \dfrac{1 – \cos (x)}{x \cos (x)} = \frac{1 – \cos (0)}{0 \cdot \cos (0)} = \frac{1-1}{0 \cdot 1} = \frac{0}{0}$$ Since this limit is (once again) in the form of  $\dfrac{0}{0}$,  it is indeterminate—we don’t yet know what it is. So we have more work to do.

Recall the second Special Limit above:

$$\lim_{x \to 0}\frac{1-\cos(x)}{x} = 0$$ We see it buried in the limit we’re given in the question, so let’s regroup the terms to see it clearly:

\begin{align*} \lim_{x \to 0} \dfrac{1 – \cos (x)}{x \cos (x)} &= \lim_{x \to 0} \left( \dfrac{1 – \cos (x)}{x}\right)\left( \frac{1}{ \cos (x)}\right) \\[12px] &= \left( \lim_{x \to 0} \dfrac{1 – \cos (x)}{x}\right)\left( \lim_{x \to 0} \frac{1}{ \cos (x)}\right) \\[12px] &= (0)\left( \frac{1}{ \cos (0)}\right) \\[12px] &= (0)\left( \frac{1}{1}\right) = 0 \quad \cmark \end{align*}
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Problems where $x \to \infty$ deserve their own topic: Limits at Infinity.
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