Question: How do I find and prove the limit of the following function using the epsilon delta definition.

$$\lim\_{x \to 0} \frac{2+\sin x}{3-\cos x}x$$

Answer: 
We know $\frac{1}{4}\leq |\frac{2+\sin x}{3-\cos x}|\leq \frac{3}{2} $. 

Then for any $\epsilon>0$, there exist $\delta=\frac{2}{3}\epsilon$ such that if $|x|<\delta$, we have 
\begin{equation}
|\frac{2+sinx}{3-cosx}x|< \frac{3}{2} \times \delta =\frac{3}{2}\times \frac{2}{3}\epsilon=\epsilon
\end{equation}

This means 
\begin{equation}
\lim\_{x\rightarrow 0} \frac{2+sinx}{3-cosx}x=0
\end{equation}
Firstly we assume the limit is 0, now we need to show:

$\forall\epsilon \gt0 ,\, \exists\delta\gt 0 :|x-0|\lt\delta\Rightarrow|\frac{2+\sin x}{3-\cos x}x-0|\lt\epsilon$ 

Now:

$|\frac{2+\sin x}{3-\cos x }x - 0 |=|\frac{2+\sin x}{3-\cos x}x|\leq\frac {3}{2}|x|\lt\frac {3}{2}\delta$ 

So we pick $\delta=\frac { 2 } { 3 } \epsilon$ then if $|x|\lt \delta$

$\Rightarrow |\frac{2+\sin x}{3-\cos x }x - 0 |\lt\frac {3}{2}\delta=\epsilon$

Thus the limit is 0