  ### Home > APCALC > Chapter 3 > Lesson 3.3.4 > Problem3-140

3-140.

Evaluate each limit. If the limit does not exist due to a vertical asymptote, then add an approach statement stating if $y$ is approaching negative or positive infinity.

1. $\lim\limits _ { x \rightarrow - 2 } ( 2 x ^ { 2 } - 6 x + 5 ) ^ { 2 }$

Evaluate.

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

As in all limits in which $x\rightarrow \infty$ or $x\rightarrow -\infty$, compare only the highest power in the numerator and denominator.

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

Careful! This one can be deceptive. Just because this is a one-sided limit, does not mean that the limit $\rightarrow \infty$ or $−\infty$. That will only be the case if there is a zero in the denominator that cannot be cancelled out.

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

This limit is asking you to determine whether the function approaches its vertical asymptote at $x = 2$ in a positive or negative direction when looked at from the left.

Choose a value a little to the left of $v$ and evaluate. You only need to determine if the limit is positive or negative.

$\lim\limits_{x\rightarrow 1.9}\frac{1.9+2}{(1.9-1)(1.9-2)}=\lim\limits_{x\rightarrow 1.9}\frac{+}{(+)(-)}$

$=$ negative value

Limit does not exist, but $y\rightarrow −\infty$.