Question

$\underset{x\to 0}{lim}\frac{e^{\sin x}-\sin x-1}{x^2}$ is equal to

• A. $\frac{1}{3}$
• B. $1$
• C. $e$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{e^{\sin x}-\sin x-1}{x^2}=\underset{x\to 0}{lim}\frac{(1+\frac{\sin x}{1!}+\frac{{\sin }^{2}x}{2!}+\frac{{\sin }^{3}x}{3!}+...)-\sin x-1}{x^2}=\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2}(\frac{1}{2!}+\frac{\sin x}{3!}+...)=\frac{1}{2}$