Use series to evaluate the limit.
-$$\lim _ { x \rightarrow 0 } \frac { \sin ^ { - 1 } 3 x - \sin 3 x } { \tan ^ { - 1 } 3 x - \tan 3 x }$$
A) $- \frac { 1 } { 2 }$
B) $- \frac { 2 } { 3 }$
C) $- \frac { 1 } { 3 }$
D) $\frac { 1 } { 3 }$

Question

Quizplus · Accepted Answer

The correct answer is A because the limit is equal to -1/2.

Use Series to Evaluate the Limit lim⁡x→0sin⁡−13x−sin⁡3xtan⁡−13x−tan⁡3x\lim _ { x \rightarrow 0 } \frac { \sin ^ { - 1 } 3 x - \sin 3 x } { \tan ^ { - 1 } 3 x - \tan 3 x }x→0lim​tan−13x−tan3xsin−13x−sin3x​

Use Series to Evaluate the Limit $\lim _ { x \rightarrow 0 } \frac { \sin ^ { - 1 } 3 x - \sin 3 x } { \tan ^ { - 1 } 3 x - \tan 3 x }$