Limits; factoring: $\lim_{x \to 2} \frac {\sqrt {x+2}-2}{x-2}$

Question

$$ \lim\limits_{x \to 2} \frac { \sqrt { x + 2 } - 2 } { x - 2 } $$

I'm having so much trouble with this one

Answer 1

If we multiply this fraction by \begin{equation} \dfrac{\sqrt{x+2}+2}{\sqrt{x+2}+2} \end{equation} (just another way to write 1), we have \begin{equation} \lim_{x\to 2}\dfrac{(x+2)-4}{(x-2)(\sqrt{x+2}+2)} = \lim_{x\to 2}\dfrac{x-2}{(x-2)(\sqrt{x+2}+2)} = \lim_{x\to 2}\dfrac{1}{\sqrt{x+2}+2}. \end{equation} Do you see how this would help?

Answer 2

HINT:

Another approach: let $$\sqrt{x+2}-2=y\implies x=(y+2)^2-2$$

As $x\to2, y\to0$

Finally as $y\to0,y\ne0$ so can be cancelled safely.

Answer 3

You may notice that if $x+2>0$ then $$x-2= (x+2)-4 = (\sqrt{x+2})^2-2^2=(\sqrt{x+2}-2)(\sqrt{x+2}+2).$$

It is basically noticing an instance of the formula $a^2-b^2=(a-b)(a+b)$.

After this is should be rather straightforward how to continue.

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You might also have a look at answers to this question, which is kind of similar: How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$?

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It is also worth mentioning that directly from the definition of derivative you can see that this is $f'(2)$ for $f(x)=\sqrt{x+2}$. $$f'(2)=\lim\limits_{x\to2}\frac{f(x)-f(2)}{x-2} = \lim\limits_{x\to2}\frac{\sqrt{x+2}-2}{x-2}.$$ But I guess that problems like this are typically often assigned before you learn about derivatives. (And, moreover, you need to be able to find at least one similar limit when you are proving what derivative of $\sqrt{x}$ looks like. Once you prove that, you can of course use it, but otherwise this would lead to circular reasoning.)