Answer
$2$
Work Step by Step
Solve $\lim\limits_{\theta \to 0}\dfrac{e^{\theta }-e^{-\theta }-2\theta }{\theta -\sin \theta }$
Now, $\lim\limits_{\theta \to 0}\dfrac{e^{\theta }-e^{-\theta }-2\theta }{\theta -\sin \theta }=\lim\limits_{\theta \to 0}\dfrac{(1+\theta+\dfrac{\theta^2}{2!}+\dfrac{\theta^3}{3!}+...)-(1-\theta+\dfrac{(-\theta)^2}{2!}+\dfrac{(-\theta)^3}{3!}+...)-2(1+\theta+\dfrac{\theta^2}{2!}+\dfrac{\theta^3}{3!}+...)}{\theta -(\theta-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}-..)}$
$\implies \lim\limits_{\theta \to 0}[\dfrac{2 \theta^3 (\dfrac{\theta^2}{6}+\dfrac{\theta^4}{120}-...)}{\theta^3(\dfrac{1}{6}-\dfrac{\theta^2}{120}+...)}]=2$
