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.
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.
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} &= \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.
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*}
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.
1. Remember your fundamental trig identities. For instance:
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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