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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 $$ 0<|x-0|<\delta \implies \frac{x}{2-\sin^2 x}<\varepsilon $$

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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 $\left\vert \frac{f(x)-L}{x-a}\right\vert\le B$

$$ \vert f(x)-L\vert=\vert x-a\vert\cdot\left\vert \frac{f(x)-L}{x-a}\right\vert<\frac{\epsilon}{B}\cdot B=\epsilon$$

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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*}

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Hint: $\sin^2x\le1$ for all $x$, so $2-\sin^2x\ge1$, hence

$$\left|x\over2-\sin^2x\right|\le|x|$$

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