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2-118.

Evaluate the following limits.

$\lim\limits_ { x \rightarrow 3 } \frac { x ^ { 2 } - 9 } { x - 3 }$
Hint 1(a):
Factor the numerator.
Hint 2(a):
Simplify.
Hint 3(a):
A limit exists at a hole. What is the $y$ -value of the hole?

$\lim\limits_ { x \rightarrow \infty } \sqrt { \frac { 16 x ^ { 2 } - 1 } { 4 x ^ { 2 } - 1 } }$
Hint (b):
Notice that this is a limit $x→∞$ . Compare the greatest powers of the numerator and denominator.
More Help (b):
$\lim\limits_{x\rightarrow \infty }\sqrt{\frac{16x^{2}}{4x^{2}}}=\lim\limits_{x\rightarrow \infty }\frac{4x}{2x}=$

$\lim\limits_ { x \rightarrow \infty } \frac { 2 ^ { - x } } { 2 ^ { x } }$
Hint (c):
Recall that $2^{-x}=\frac{1}{2^x}$ .
More Help (c):
$\lim\limits_{x\rightarrow \infty }\frac{1}{2^{2x}}=\lim\limits_{x\rightarrow \infty }\frac{1}{4^x}=$

$\lim\limits_ { x \rightarrow \infty } \frac { 2 x ^ { 3 } - x - 1 } { 12 - x - x ^ { 2 } }$
Hint (d):
This is another limit $x→∞$ . Compare the terms with the highest powers in the numerator and denominator.
More Help (d):
$\lim\limits_{x\rightarrow \infty }\frac{2x^3}{-x^{2}}=\lim\limits_{x\rightarrow \infty }\frac{2x}{-1}=$

$\lim\limits_ { x \rightarrow \infty } \operatorname { sin } x$
Hint (e):
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.

$\lim\limits_ { x \rightarrow \infty } \frac { \operatorname { sin } x } { x }$
Hint (f):
Consider the numerator and the denominator separately.
More Help (f):
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.

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