Proving a limit using epsilon delta definition. How can I come about and prove this limit using epsilon delta? I'm kind of guessing I have to bound 2-sin^2(x), but i'm not quite sure...
Help would be much appreciated 

$$
\lim_{x\to 0}\frac{x}{2-\sin^2 x} = 0
$$
  Given $\varepsilon > 0$, we look for $\delta$ such that
  $$
0<|x-0|<\delta \implies \left|\frac{x}{2-\sin^2 x} - 0\right|<\varepsilon
$$
  or
  $$
0<|x-0|<\delta \implies \frac{x}{2-\sin^2 x}<\varepsilon
$$

 A: Let $\epsilon>0$ and $\vert x-0\vert<\epsilon$. Let $\delta=\epsilon$.
\begin{eqnarray}
1&\ge&\sin^2x\\
2&\ge&1+\sin^2x\\
2-\sin^2x&\ge&1\\
\frac{1}{2-\sin^2x}&\le&1\\
\left\vert\frac{x}{2-\sin^2x}-0\right\vert&\le&\left\vert x\right\vert<\epsilon=\delta
\end{eqnarray}
ADDENDUM: When trying to prove that 
$$ \lim_{x\to a} f(x)=L$$
by the $\epsilon,\delta$ method, first try to find an upper bound $B$ satisfying
$$  \left\vert \frac{f(x)-L}{x-a}\right\vert\le B$$
on some interval $(a-r,a+r)$ for which $\left\vert \frac{f(x)-L}{x-a}\right\vert$ is bounded.
[Note: It is not always the case that such a $B$ exists.]
If you are able to find such a bound $B$ on some interval $(a-r,a+r)$, then let $\epsilon>0$ and let $\delta=\min\left\{r,\frac{\epsilon}{B}\right\}$.
It follows that if $\vert x-a\vert<\delta$, then $\left\vert \frac{f(x)-L}{x-a}\right\vert\le B$
$$ \vert f(x)-L\vert=\vert x-a\vert\cdot\left\vert \frac{f(x)-L}{x-a}\right\vert<\frac{\epsilon}{B}\cdot B=\epsilon$$
A: Hint: From $|\sin(x)|\leq|x|$ for all $x\in\mathbb R$ it follows that
\begin{align*}
\left|\frac{x}{2-\sin^2(x)}\right|\leq\frac{|x|}{2-|\sin(x)|^2}\leq\frac{|x|}{2-|x|^2}.
\end{align*}
A: Hint:  $\sin^2x\le1$ for all $x$, so $2-\sin^2x\ge1$, hence
$$\left|x\over2-\sin^2x\right|\le|x|$$
