Home > APCALC > 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 evaluate the limit as the function approaches $∞$, look at the end behavior of the function.

Determine 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 a $y=x^2$ graph, we know that the limit as the function approaches ∞ does not exist.

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 are different end-behavior limits in either direction.

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

This limit has an indeterminate form. Use l'Hôpital's Rule.

$\lim\limits_{x\to0}\frac{\sin(2x)-8x^2}{4x(x-10)^3}=\lim\limits_{x\to0}\frac{2\cos(2x)-16x}{4(x-10)^3+12x(x-10)^2}$

Now try evaluating the limit.

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

This limit also has an indeterminate form. Use l'Hôpital's Rule.