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Using the $\epsilon-\delta$ definition of the limit, evaluate $\lim_{x \to 0} f(x)$, where $f(x) = \dfrac{x}{1+\sin^2{x}}$.

Attempt

We need to show that $$\forall \epsilon, \exists \delta \quad 0 < |x-a| < \delta \quad \implies \quad \left | \dfrac{x}{1+\sin^2{x}}-\left ( \dfrac{a}{1+\sin^2{a}} \right) \right| < \epsilon.$$

The problem here is that I can't really factor out $|x-a|$ easily from that so I am lost how to proceed next. We can substitute in $a = 0$ to get $\left| \dfrac{x}{1+\sin^2{x}} \right| < \epsilon$ but then how do we relate delta?

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  • $\begingroup$ Notice that $1+\sin^2(x)≥1$, so $1/(1+\sin^2(x))≤1$ for all $x$. If $|x|<\delta$ how can you combine these two things and choose $\delta$ to get a helpful expression? Aside from that, $\epsilon-\delta$ usually refers to the definition of continuity and not to the evaluation of a limit. I believe your exercise does not want you to evaluate a limit but rather show continuity of a function at zero. $\endgroup$ – s.harp Mar 6 '16 at 20:21
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Note that since $1+\sin^2 x \geq 1$, we have $$ \left|\frac{x}{1 + \sin^2 x}\right| \leq \left|\frac{x}{1}\right| = |x| $$ Clearly, if $|x|<\epsilon$, then $\left|\frac{x}{1 + \sin^2 x}\right| < \epsilon$. Therefore, it suffices to pick $\delta = \epsilon$.

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  • $\begingroup$ What if $|x| \not < \epsilon$? $\endgroup$ – user19405892 Mar 6 '16 at 20:28
  • $\begingroup$ @user19405892 But it is, if you choose $\delta = \epsilon$. We are, after all, only interested in $|x|<\delta$. $\endgroup$ – Arthur Mar 6 '16 at 20:30

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