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.