This is a one-sided limit because the domain of $y=\sqrt{x-4}-5$ begins at $x=4$. Consequently, the limit exists from the right, but not from the left.

Factor first. If you can 'cancel out' the denominator, then the graph has a hole (not a vertical asymptote) and the limit exists.

Evaluate. The denominator will not equal zero. So the limit and the actual value agree.

Visualize the graph of $y=\frac{1}{x-1}$. Both graphs have the same end behavior (horizontal asymptote).

For limits in which $x→∞$ or $x→−∞$, we are looking for end behavior. For example, is there a horizontal asymptote?

Since the numerator has a higher power, $x^2$, than the denominator, $x$, there is no horizontal asymptote. Thus the limit goes to either $+∞$ or $−∞$.

The limit goes to $−∞$ because:
$\lim\limits_{x\rightarrow -\infty }\frac{x^{2}}{x}=\lim\limits_{x\rightarrow -\infty }x=-\infty$

Refer to hint in part (e).