Use series to evaluate the limit.
-$$\lim _ { x \rightarrow 0 } \frac { \sin 10 x - \sin 20 x } { x }$$
A) -10
B) -1
C) 10
D) 1

Question

Quizplus · Accepted Answer

Using the Taylor series expansion for sine, $\sin x \approx x - \frac{x^3}{6} + \cdots$, we have $\sin 10x \approx 10x$ and $\sin 20x \approx 20x$ for small $x$. Thus, $\sin 10x - \sin 20x \approx 10x - 20x = -10x$. Therefore, $\lim_{x \to 0} \frac{\sin 10x - \sin 20x}{x} = \lim_{x \to 0} \frac{-10x}{x} = -10$.

Use Series to Evaluate the Limit lim⁡x→0sin⁡10x−sin⁡20xx\lim _ { x \rightarrow 0 } \frac { \sin 10 x - \sin 20 x } { x }x→0lim​xsin10x−sin20x​

Use Series to Evaluate the Limit $\lim _ { x \rightarrow 0 } \frac { \sin 10 x - \sin 20 x } { x }$