Use series to evaluate the limit correct to three decimal places. $\lim _ { x \rightarrow 0 } \frac { 7 x - \tan ^ { - 1 } 7 x } { x ^ { 3 } }$ Select the correct answer.
A) 118.933
B) 114.133
C) 34.3233
D) 114.333
E) 115.933

Question

Quizplus · Accepted Answer

The series expansion for $\tan^{-1}(x)$ around $x=0$ is $x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$. Substituting $7x$ for $x$, we get $\tan^{-1}(7x) = 7x - \frac{(7x)^3}{3} + \cdots$. The expression becomes $\frac{7x - (7x - \frac{343x^3}{3})}{x^3} = \frac{\frac{343x^3}{3}}{x^3} = \frac{343}{3} = 114.333$.

Use Series to Evaluate the Limit Correct to Three Decimal lim⁡x→07x−tan⁡−17xx3\lim _ { x \rightarrow 0 } \frac { 7 x - \tan ^ { - 1 } 7 x } { x ^ { 3 } }limx→0​x37x−tan−17x​

Use Series to Evaluate the Limit Correct to Three Decimal $\lim _ { x \rightarrow 0 } \frac { 7 x - \tan ^ { - 1 } 7 x } { x ^ { 3 } }$