This is a one-sided limit because the domain of

$y=\sqrt{x-4}-5 \text{ begins at }x=4.$

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.

$\text{Visualize the graph of }y=\frac{1}{x-1}.$

$\text{It is a transformation of }y=\frac{1}{x}.$

Both graphs have the same end behavior (horizontal asymptote).

For limits in which $x \rightarrow \infty$ or $x \rightarrow -\infty$, 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 $+\infty$ or $-\infty$.

The limit goes to $-\infty$ because:

$\lim\limits _{x\rightarrow -\infty }\frac{x^{2}}{x}=\lim\limits _{x\rightarrow -\infty }x=-\infty$

Refer to hint in part (e).