### Home > CALC > Chapter 2 > Lesson 2.3.2 > Problem2-118

2-118.

Evaluate the following limits.

1. $\lim\limits_ { x \rightarrow 3 } \frac { x ^ { 2 } - 9 } { x - 3 }$

Factor the numerator.

Simplify.

A limit exists at a hole. What is the $y$-value of the hole?

1. $\lim\limits_ { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } - 1 } { 4 x ^ { 2 } - 1 } }$

Notice that this is a limit $x→∞$. Compare the greatest powers of the numerator and denominator.

$\lim\limits_{x\rightarrow \infty }\sqrt{\frac{16x^{2}}{4x^{2}}}=\lim\limits_{x\rightarrow \infty }\frac{4x}{2x}=$

1. $\lim\limits_ { x \rightarrow \infty } \frac { 2 ^ { - x } } { 2 ^ { x } }$

Recall that $2^{-x}=\frac{1}{2^x}$.

$\lim\limits_{x\rightarrow \infty }\frac{1}{2^{2x}}=\lim\limits_{x\rightarrow \infty }\frac{1}{4^x}=$

1. $\lim\limits_ { x \rightarrow \infty } \frac { 2 x ^ { 3 } - x - 1 } { 12 - x - x ^ { 2 } }$

This is another limit $x→∞$. Compare the terms with the highest powers in the numerator and denominator.

$\lim\limits_{x\rightarrow \infty }\frac{2x^3}{-x^{2}}=\lim\limits_{x\rightarrow \infty }\frac{2x}{-1}=$

1. $\lim\limits_ { x \rightarrow \infty } \operatorname { sin } x$

Think about the graph of $y = \operatorname{sin}x$. It oscillates as $x→∞$. This means that it does not approaches a finite value AND does not approach $±∞$. When a function oscillates, we say the limit does not exist.

1. $\lim\limits_ { x \rightarrow \infty } \frac { \operatorname { sin } x } { x }$

Consider the numerator and the denominator separately.

As $x→∞$, $y =\operatorname{sin}x$ oscillates between $y = −1$ and $y=1$.
As $x→∞$, $y = x$ approaches $∞$.
So this ratio has small values on the top and infinity on the bottom.