Another tactic you *must* have in your toolbox is using conjugates to find a limit. Let’s look at examples and then you can practice using our problems with complete solutions.

Specifically, if you try Substitution and obtain the fraction $\dfrac{0}{0}$ *and* the expression has a square root in it, then use conjugates just like you practiced in Algebra. That is, multiply both the numerator and the denominator by the conjugate of the term with the square root. (Recall that $\left(\sqrt{a} + \sqrt{b} \right)$ and $\left(\sqrt{a} -\sqrt{b} \right)$ are conjugates of each other: one has a plus-sign between the terms, while the other has a minus-sign.)

You may find that this requires introducing a square root in the denominator where there wasn’t one before. If this move runs counter to what you’ve learned in earlier math classes, that “there should never be a square root in the denominator,” then we have to tell you now: that’s a silly grade-school rule that has no place in your ongoing mathematical development, as you’ll see as you work through the problems below.

An example best illustrates the approach.

Example: Use Conjugates to Find a Limit

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

*Solution.*

**Step 1. Try Substitution:**

\[\lim_{x \to 0}\frac{\sqrt{x+5} – \sqrt{5}}{x} \overset{?}{=} \frac{\sqrt{0+5}-\sqrt{5}}{0} = \frac{0}{0}\]
Since the limit is in the form $\dfrac{0}{0}$ , substitution gave an indeterminate result — we have more work to do.

**Step 2. Multiply by conjugate and simplify:**

Let’s use algebra to get rid of the square roots: multiply both the numerator and denominator by the conjugate of the numerator, $\sqrt{x+5} + \sqrt{5}$. We can do this since we are multiplying the original expression by $1= \dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}},$ which doesn’t change the value of anything.

\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\left[ \sqrt{x+5} +

\sqrt{5}\right]} \\[8px]
&= \lim_{x \to 0}\dfrac{(x+5) – 5}{x[\sqrt{x+5} + \sqrt{5}]} \quad\color{blue}{\text{[Note we have not multiplied the denominator terms…]}}\\[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}]} \quad\color{blue}{\text{[… because then the $x$ then cancels nicely here.]}}\\[8px]
&= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\[8px]
\end{align*}

**Step 3. Now easy Substitution to finish:**\begin{align*}

\phantom{\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x}} &\phantom{= \lim_{x \to 0}\dfrac{(x+5) – 5}{x[\sqrt{x+5} + \sqrt{5}]} \text{[Note we have not multiplied the denominator terms]}}\\

&=\dfrac{1}{\sqrt{0+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} \quad \cmark

\end{align*}

This approach works for essentially the same reason the factoring tactic works: the functions $\dfrac{\sqrt{x+5} – \sqrt{5}}{x}$ and $\dfrac{1}{\sqrt{x+5} + \sqrt{5}}$ are the same, *except* that the original function is *not* defined at $x=0,$ whereas the rewritten function *is*.

Hence their limits are the same as $x \to 0$, and so $\displaystyle{\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} = \lim_{x\to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} }$.

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.

The Example illustrates the key steps to using conjugates to find a limit, which mirror quite closely the steps we used for Factoring:

PROBLEM-SOLVING TACTIC: Use conjugates to find a limit

**Try Substitution.**As we’ve seen, if it works immediately as a tactic you’re done. If, by contrast, you obtain $\dfrac{0}{0}$ with a square-root in the numerator and/or denominator, then…**Multiply by the conjugate and simplify.****Use Substitution to finish.**

As always, the best way to get the tactic down for yourself is to practice using it until it becomes routine. That’s even more true here than in factoring on the preceding screen, since the algebra tends to get a bit messier with these, given all those square-root signs floating around. As in all things, practice lets you become more comfortable with “the mess,” so you won’t make a silly mistake when under exam pressure.

Use Conjugates to Find a Limit: Scaffolded Exercise #1

Find $\displaystyle{\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}}$.

**Solution.**

**Step 1. Try Substitution:**

Use Conjugates to Find a Limit: Scaffolded Exercise #2

Find $\displaystyle{\lim_{x \to 16} \frac{\sqrt{x}-4}{x-16} }.$

**Solution.**

**Step 1. First try substitution:**

Practice Problem #1

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}

Practice Problem #2

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}

Practice Problem #3

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}

Practice Problem #4

Find $\displaystyle{\lim_{x \to 6}\dfrac{\sqrt{x-2}-2}{x-2} }.$
\begin{array}{lllll} \text{(A) }0 && \text{(B) }\dfrac{1}{4} && \text{(C) }-\dfrac{1}{4} && \text{(D) DNE} && \text{(E) None of these} \end{array}

Practice Problem #5

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}

Practice Problem #6

Find $\displaystyle{ \lim_{x \to 0} \left(\frac{1}{x \sqrt{1+x}} - \frac{1}{x} \right) }$.

\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}

Open for hint.

First put both terms over a common denominator. Then proceed as usual.

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- If when you try Substitution you obtain $\dfrac{0}{0}$ and you have a square root in the numerator and/or denominator, then multiply by the conjugate of the square-root term divided by itself. After simplifying, Substitution will work.

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