Prove the limit doesn't exist using basic Calculus I was given this problem by a friend:
$$
\def\limit{\lim_{x\to5}}
\def\top{\sqrt{x}-2}
g(x) = \frac{\top}{x-5}\\
\limit g(x) = \quad?
$$
This caught me by surprise, because I can't remember how to do this problem with basic Calculus. Intuitively, the limit doesn't exist since $\limit(\top) \ne 0$ but $\limit(x-5) = 0$. And the limit doesn't even approach either infinity since the bottom is an odd-power polynomial.
However, how can I prove this by using basic Calculus that a new Calculus student would understand?

I tried to do this as follows, but it seems overly complicated:
I spent some time and found this function:
$$
f(x)=\frac1{\sqrt{x-5}}\\
f(x) < \frac\top{x-5}\\
\text{when }5 < x < \frac{81}{16}
$$
And we know that $\lim_{x\to5^+} f(x) = \infty$, so therefore,
$$\lim_{x\to5^+} g(x) = \infty$$
However, $f(x)$ doesn't work for the LH limit. However:
$$
g(x) > 0 \quad\text{if}\quad x > 5\\
g(x) < 0 \quad\text{if}\quad 4 < x < 5
$$
So that means that $\lim_{x\to5^-} g(x) < 0 \ne \infty$ so the limit does not exist, and we can't even say that the limit is one of the infinities.
Isn't there an easier way to do this (assuming graphing isn't allowed)?
 A: Probably the best way to do this is your original argument:
If $\displaystyle\lim_{x\to c} f(x)\ne 0$ and $\displaystyle\lim_{x\to c} g(x)=0$, then it follows that
$\displaystyle\lim_{x\to c} \frac{f(x)}{g(x)}$ does not exist, since
$\displaystyle\lim_{x\to c} \frac{f(x)}{g(x)}=L\implies\lim_{x\to c}f(x)=\lim_{x\to c} \left(\frac{f(x)}{g(x)}\right)\big(g(x)\big)=L\cdot0=0$, which gives a contradiction.
A: Hint: $$\frac{\sqrt x - 2}{x-5}=\frac{(\sqrt x -\sqrt5)+(\sqrt5- 2)}{x-5}=\frac{\sqrt x -\sqrt5}{x-5}+\frac{\sqrt 5 - 2}{x-5}=\frac{\sqrt x -\sqrt5}{(\sqrt x +\sqrt 5)(\sqrt x -\sqrt 5)}+\frac{\sqrt 5 - 2}{x-5}=\frac 1 {\sqrt x+\sqrt 5}+ \frac{\sqrt 5 - 2}{x-5}$$

This kind of trick is really useful and happens often. It is easiest to illustrate how to use it on a simple example, $\frac{x}{x+1}$. The aim is to find something where adding and subtracting it will enable us to cancel the denominator. In this case, we notice that adding and subtracting $1$ will do the trick: $$\frac{x}{x+1}=\frac{(x+1) -1}{x+1}=1-\frac{1}{x+1}$$In the above case, spotting that $x-5 = (\sqrt x +\sqrt 5)(\sqrt x -\sqrt 5)$ leads to the choice of adding and subtracting $\sqrt 5$ and lose one of the factors in the denominator.
A: Distinguish between the two cases:
$\lim_{x  \to 5^{\color{blue}{-}}} \frac{\sqrt x -2}{x-5}$ and 
$\lim_{x  \to 5^{\color{blue}{+}}} \frac{\sqrt x -2}{x-5}$
These limits can be easily evaluated. When $x \to 5^{\color{blue}{-}}$, the denominator goes to 0 and is negative while the numerator doesn't/isn't, leading the fraction to go to $-\infty$. Similarly for the other side.
