### Home > APCALC > Chapter 2 > Lesson 2.4.1 > Problem2-142

2-142.

Determine the values of the following limits. If the limit does not exist, indicate why not.

1. $\lim\limits _ { x \rightarrow 4 ^ { + } } ( \sqrt { x - 4 } - 5 )$

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.

1. $\lim\limits _{ x \rightarrow - 2 } \large\frac { x ^ { 2 } - 4 x - 12 } { x + 2 }$

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

1. $\lim\limits _ { x \rightarrow 6 } \large\frac { x ^ { 2 } - 4 x - 12 } { x + 2 }$

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

1. $\lim\limits _ { x \rightarrow \infty } \large\frac { 1 } { x - 1 }$

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

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

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

1. $\lim\limits _ { x \rightarrow - \infty } \large\frac { x ^ { 2 } + 6 x - 7 } { x }$

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:

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

1. $\lim\limits _ { x \rightarrow \infty } \large\frac { x ^ { 2 } - 7 x - 10 } { x ^ { 2 } }$

Refer to hint in part (e).