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


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?

  • $\begingroup$ 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

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 $\delta = \epsilon$.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.