Answer
$-1$
Work Step by Step
Here, we have $\lim\limits_{\theta \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{\theta \to 0} \dfrac{-\sin \theta}{e^\theta-1}=\dfrac{0}{0}$
Now, again apply L-Hospital's rule.
$\lim\limits_{\theta \to 0} \dfrac{-\cos \theta}{e^\theta}=\dfrac{-\cos 0}{e^{0}}$
or, $\dfrac{-1}{1}=-1$
