Limits with Radicals

(Videos) Calculating a Limit by Multiplying by a Conjugate

by PatrickJMT

Here's a video by patrickJMT showing you how to find the limit of a function using the conjugate technique.

He shows how to use this technique by looking at an example. $$\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}$$ Notice what happens when we plug in 7. $$\begin{align} \frac{\sqrt{x+2}-3}{x-7}&=\frac{\sqrt{\color{red}{(7)}+2}-3}{\color{red}{(7)}-7}\\ &=\frac{3-3}{7-7}\\ &=\frac00 \end{align}$$ We have to do something to this in order to change it. The key things to spot are that there's a radical and two terms in the numerator. A common trick when we have a radical is to multiply by the conjugate.

Remember that if you have $a\pm b$ the conjugate is the same thing but the sign is reversed: $a\mp b$.

So let's multiply the top and bottom by the conjugate of the numerator. $$\begin{align} &\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}\\ &=\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}\color{red}{\cdot\frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}}\\ &=\lim_{x\rightarrow7}\frac{(x+2)-9}{(x-7)(\sqrt{x+2}+3)}\\ &=\lim_{x\rightarrow7}\frac{x-7}{(x-7)(\sqrt{x+2}+3)}\\ &=\lim_{x\rightarrow7}\frac{1}{\sqrt{x+2}+3} \end{align}$$ Now we can plug in $x=7$. $$\begin{align} \frac{1}{\sqrt{x+2}+3}&=\frac{1}{\sqrt{\color{red}{(7)}+2}+3}\\ &=\frac1{3+3}\\ &=\frac16 \end{align}$$

So $$\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}=\frac16$$