Answer
$2$
Work Step by Step
We know that the Taylor Series for $e^x$ and $\sin x$ is defined as:
$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...$ and $ \sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-...$
Now, $\lim\limits_{\theta \to 0}\dfrac{e^{\theta }-e^{-\theta }-2\theta }{\theta -\sin \theta }=\lim\limits_{\theta \to 0}\dfrac{(1+\theta+\dfrac{(\theta)^2}{2!}+\dfrac{(\theta)^3}{3!}+...)-(1-\theta+\dfrac{(-\theta)^2}{2!}+\dfrac{(-\theta)^3}{3!}+...)-2(1+\theta+\dfrac{(\theta)^2}{2!}+\dfrac{(\theta)^3}{3!}+...)}{\theta -(\theta-\dfrac{(\theta)^3}{3!}+\dfrac{(\theta)^5}{5!}-..).}$
or, $=\lim\limits_{\theta \to 0}[\dfrac{2 (\theta)^3 (\dfrac{(\theta)^2}{3!}+\dfrac{(\theta)^4}{5!}-...)}{\theta^3(\dfrac{1}{3!}-\dfrac{(\theta)^2}{5!}+...)}]$
or, $\dfrac{2(\dfrac{1}{6}+0+...)}{(\dfrac{1}{6}-0+...)}=2$
