### Home > CALC > Chapter 6 > Lesson 6.1.1 > Problem6-13

6-13.

Evaluate each limit.

1. $\lim\limits_ { x \rightarrow \infty } \frac { x ^ { 5 } - 4 x ^ { 3 } + 3 x ^ { 2 } - 1 } { x ^ { 3 } + 8 x ^ { 2 } + 21 x - 5 }$

To find the limit as the function approaches $∞$, look at the end behavior of the function.

Find the end behavior by looking at only the highest-power terms in the denominator and numerator.
$=\lim\limits_{x\rightarrow \infty }\frac{x^{5}}{x^{3}}=\lim\limits_{x\rightarrow \infty }x^{2}=$

Because the function's end behavior follows the trend of an $x^2$ graph, we know that the limit as the function approaches $∞$ is also $∞$.

1. $\lim\limits_ { x \rightarrow - \infty } \frac { x ^ { 5 } - 4 x ^ { 3 } + 3 x ^ { 2 } - 1 } { x ^ { 3 } + 8 x ^ { 2 } + 21 x - 5 }$

This is ALMOST the same problem as part (a) but, evaluate the limit as $x→−∞$ instead of $∞$.
Note: sometimes there is are different end-behavior limits in each direction.

1. $\lim\limits_ { x \rightarrow \infty } \frac { 10 x ^ { 4 } + 5 x ^ { 3 } + x ^ { 2 } - 1 } { x ^ { 5 } + x ^ { 2 } - 12 x - 5 }$

Refer to part (a). Remember to include $10$ as part of the highest power for the numerator!

1. $\lim\limits_ { x \rightarrow \infty } \frac { x ^ { 4 } + 7 x ^ { 3 } - 5 x ^ { 2 } - 8 } { 6 x ^ { 4 } + 3 x ^ { 3 } - 9 x + 2 }$

See part (a).

$=(\text{show steps})=\frac{1}{6}.$
Note: this means that there is a horizontal asymptote of $y=\frac{1}{6}.$

1. $\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { sin } ( 2 x ) - 8 x ^ { 2 } } { 4 x ( x - 10 ) ^ { 3 } }$

1. $\lim\limits_ { x \rightarrow 2 } \frac { 8 ( x - 2 ) ^ { 2 } } { x ^ { 2 } - 5 x + 6 }$