Find the limit (if it exists). $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x }$
A) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = - 8$
B) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = 72$
C) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = 64$
D) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = 8$
E) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = - 64$

Question

Find the limit (if it exists).   $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x }$  
A) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = - 8$
B) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = 72$
C) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = 64$
D) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = 8$
E) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x } = - 64$

Quizplus · Accepted Answer

The limit simplifies to $\lim _ { x \rightarrow \pi } \frac { 8 } { \sin x } \cdot \tan x = \lim _ { x \rightarrow \pi } \frac { 8 \sin x } { \sin x \cos x } = \lim _ { x \rightarrow \pi } \frac { 8 } { \cos x } = -8$, since $\cos(\pi) = -1$.

Find the Limit (If It Exists) lim⁡x→π8csc⁡xcot⁡x\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x }limx→π​cotx8cscx​

Find the Limit (If It Exists) $\lim _ { x \rightarrow \pi } \frac { 8 \csc x } { \cot x }$