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Evaluating Limits

Evaluating Limits: Problems and Solutions

You probably already understand the basics of what limits are, 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 evaluate or compute a limit in homework and exam problems, especially in cases where you initially get 0 divided by 0. We’ll illustrate the six tactics you must know, and then let you practice each.

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


For a discussion of why these approaches work, please read our blog post 0 Divided by 0: Solve Limit Problems in Calculus, Part 1.

Calculus 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 Click to View Calculus Solution toggle. Substitution practice problem #3 illustrates an important point.
Substitution Practice #1
Find $\displaystyle{ \lim_{x \to -2}\left(x^3 +4\right) }$.
Click to View Calculus Solution
Answer: $-4$
\begin{align*}
\lim_{x \to -2}\left(x^3 +4\right) &= (-2)^3 +4 \\[4px] &= -8 +4 \\[4px] &= -4 \quad \cmark
\end{align*}
Substitution Practice #2
Find $\displaystyle{ \lim_{x \to 1} }\dfrac{\ln x}{x}$.
Click to View Calculus Solution
Answer: $0$
Recall $\ln (1) = 0$. Hence
\begin{align*}
\lim_{x \to 1}\dfrac{\ln x}{x} &= \dfrac{\ln (1) }{1} \\[12px] &= \frac{0}{1} \\[12px] &= 0 \quad \cmark
\end{align*}
Substitution Practice #3
Find $\displaystyle{\lim_{x \to 1} \dfrac{3}{x-1} }$.
Click to View Calculus Solution
Answer: This limit does not exist (DNE).

\[ \lim_{x \to 1} \dfrac{3}{x-1} = \dfrac{3}{1-1} = \frac{3}{0} \] Limits problem: 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:
$\boxed{\text{DNE}}\quad \cmark$

TIP
Tips icon
As the preceding problem illustrates, when you substitute the limit-value into the function, if you obtain   \(\frac{\text{a non-zero number}}{0}\text{,} \),   then you know immediately that the limit does not exist (DNE):
\[\text{If} \quad \lim_{x \to a}f(x) = \frac{\text{a non-zero number}}{0}\] \[\text{then the limit does not exist (DNE).}\]
Calculus Limits Tactic #2: Factoring
Frequently (on homework and exams, anyway) when you try Substitution, you obtain the fraction  $\dfrac{0}{0}$, 0 divided by 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:

$\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$\(=\)$\displaystyle{\lim_{x \to 2}\frac{(x+2)(x-2)}{x-2}}$
Ah, now we can cancel the problematic term:
\(=\)$\displaystyle{\lim_{x \to 2}\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}}}$
\(=\)$\displaystyle{\lim_{x \to 2} \,(x+2)}$
And now easy Substitution to finish:
\(=\)$\displaystyle{2 + 2 = 4 \quad \cmark}$

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.

Factoring Practice #1
$\displaystyle{\lim_{x \to 3} \frac{x-3}{x^2-9}}$
Click to View Calculus 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.
Limits problem: 1 over x+3 with a hole

Substitution Practice #2
Find $\displaystyle{ \lim_{x \to 1} }\frac{x^2 + 4x -5}{x^2 – 1}$.
Click to View Calculus Solution
Answer: 3


We first try substitution:
\[\lim_{x \to 1} \frac{x^2 + 4x -5}{x^2 – 1} = \frac{1 +4(1) – 5}{1-1} = \frac{0}{0}\] 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 \quad \cmark
\end{align*}
Calculus Limits Tactic #3: Conjugates
If you try Substitution and obtain the fraction $\dfrac{0}{0}$ (0 divided by 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.

Conjugates Practice #1
Find $\displaystyle{\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}}$.
Click to View Calculus Solution
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*}

Conjugates Practice #2
Find $\displaystyle{ \lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3}.}$
Click to View Calculus Solution

We first try substitution:

\[ \lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3} = \frac{\sqrt{3+1} -2}{3-3} = \frac{\sqrt{4}-2}{3-3} = \frac{0}{0} \]

Since this limit is in the form $\dfrac{0}{0}$, it is indeterminate—we don’t yet know what it is.  So let’s rationalize the expression:

\[ \begin{align*}
\lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3} &= \lim_{x \to 3}\frac{\sqrt{x+1} -2}{x-3} \cdot \frac{\sqrt{x+1} +2}{\sqrt{x+1} +2} \\[8px] &= \lim_{x \to 3} \frac{\sqrt{x+1}\sqrt{x+1}+ 2\sqrt{x+1} – 2 \sqrt{x+1} -4}{(x-3)(\sqrt{x+1}+2)} \\[8px] &= \lim_{x \to 3}\frac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}\\[8px] &= \lim_{x \to 3}\frac{x-3}{(x-3)(\sqrt{x+1}+2)} \\[8px] &= \lim_{x \to 3}\frac{1}{\sqrt{x+1}+2} \\[8px] &= \frac{1}{\sqrt{3+1}+2} \\[8px] &= \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} \\[8px] &= \frac{1}{4} \quad \cmark
\end{align*} \]

Calculus 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.
Use Algebra Practice #1
Find $\displaystyle{\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h}}$.
Click to View Calculus Solution
Answer: $-10$

As always, we first try Substitution and simply plug $h=0$ into the expression:
$$\lim_{h \to 0}\dfrac{(h-5)^2 – 25}{h} = \dfrac{(0-5)^2 -25}{0} = \dfrac{25-25}{0} = \dfrac{0}{0}$$
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} \\[12px] &= \lim_{h \to 0}\dfrac{h^2 – 10h}{h} \\[12px] &= \lim_{h \to 0}\dfrac{h(h – 10)}{h} \\[12px] &= \lim_{h \to 0}\dfrac{\cancel{h}(h-10)}{\cancel{h}} \\[12px] &= \lim_{h \to 0}(h – 10) \\[12px] &= 0 – 10 = -10 \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.

Calculus 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. It’s always available via the “Reference” link at the very top of every page.

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.
Limits: Trig Practice #1
Find $\displaystyle{\lim_{x \to 0}\dfrac{1- \cos(x) }{\sin^2(x)} }$.
Click to View Calculus Solution
Answer: $\dfrac{1}{2}$

As always, we first try Substitution:
$$\lim_{x \to 0}\dfrac{1- \cos(x)}{\sin^2(x)} = \frac{1- \cos(0)}{\sin^2(0)} = \frac{1-1}{0} = \frac{0}{0}$$
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 above, which we can rewrite as
$$ \sin^2(x) = 1 – \cos^2(x) $$
Hence

$\displaystyle{\lim_{x \to 0}\frac{1- \cos(x)}{\sin^2(x)}}$\(=\)$\displaystyle{\lim_{x \to 0}\frac{1- \cos(x)}{1 – \cos^2(x)} }$
We can factor the denominator, so let’s do it
and hope that something cancels:
\(=\)$\displaystyle{\lim_{x \to 0}\frac{1- \cos(x)}{(1 – \cos(x))(1 + \cos(x))}}$
Yep, we can cancel some terms!
\(=\)$\displaystyle{\lim_{x \to 0}\frac{\cancel{1- \cos(x)}}{\cancel{(1 – \cos(x))}(1 + \cos(x))}}$
\(=\)$\displaystyle{ \lim_{x \to 0} \frac{1}{1 + \cos(x)}}$
And now Substitution to finish:
\(=\)$\displaystyle{\frac{1}{1 + \cos(0)} }$
\(=\)$\displaystyle{\frac{1}{1+1} = \frac{1}{2} \quad \cmark}$
Limits: Trig Practice #2
Find $\displaystyle{\lim_{x \to 0}\dfrac{\sin(5x)}{x} }$.
Click to View Calculus 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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