I want to evaluate

$$\lim _{x\to \pi }\left(\frac{\sin x}{\pi ^2-x^2}\right)$$

without using L'Hopital's rule or Taylor series. My thinking process was something like this: in order to get rid of the undefined state, I need to go from $\sin x$ to $\cos x$. I tried this substitution: $t = \frac{\pi}{2}-x$ which gets me to $\cos t$, but I get:

$$\lim _{x\to -\frac{\pi}{2} }\left(\frac{\cos t}{\pi ^2-(\pi^2-t)}\right)$$

which also evaluates to $0/0$.

Then I tried this:

$\lim _{x\to \pi }\left(\frac{\sin x}{(\pi-x)(\pi+x)}\right) = \lim _{x\to \pi }\left(\frac{\sin x}{(\pi-x)2\pi}\right)$

which also gets me nowhere. The fact that $\sin(\pi-x) = \sin x$ also doesn't help because the substitution would lead to $\sin 0 = 0$.

The correct solution is $\frac{1}{2\pi}$

Can somebody help me?


hint: $\sin x = \sin(\pi - x)$, and $\pi^2 - x^2 = (\pi -x)(\pi + x)$.

Also use $\dfrac{\sin(\pi - x)}{\pi - x} \to 1$ when $x \to \pi$.

  • $\begingroup$ Oh, thank you! I was so stupid not to have noticed it! $\endgroup$
    – Quant
    Apr 15 '16 at 19:04
  • $\begingroup$ Nice. Its a learning experience... $\endgroup$
    – DeepSea
    Apr 15 '16 at 19:04
  • $\begingroup$ I was just wondering that isn't "$\dfrac{\sin(\pi - x)}{\pi - x} \to 1$ when $x \to \pi$" considered as L'Hopital's rule? $\endgroup$
    – IgNite
    Apr 15 '16 at 19:43
  • 1
    $\begingroup$ @Ignite That limit can be shown using inequalities from elementary geometry along with the squeeze theorem. I've posted an answer that walks through that development. -Mark $\endgroup$
    – Mark Viola
    Apr 15 '16 at 20:17

Note that from elementary geometry, the sine function is bounded as

$$|\theta \cos(\theta)|\le |\sin(\theta)|\le |\theta| \tag 1$$

for $0\le |\theta|\le \pi/2$. Letting $\theta =x-\pi$ in $(1)$ yields

$$|(x-\pi)\cos(x-\pi)|\le |\sin(x-\pi)|\le |x-\pi| \tag 2$$

for $0\le |x-\pi|\le \pi/2$. For $x\ne \pi$, we find upon dividing $(2)$ by $|x-\pi|$

$$\cos(x-\pi)\le \frac{\sin(x-\pi)}{x-\pi}\le 1 \tag 3$$

for $0 < |x-\pi|\le \pi/2$

Finally, dividing $(3)$ by $x+\pi$ reveals

$$\frac{\cos(x-\pi)}{x+\pi}\le \frac{\sin(x-\pi)}{x^2-\pi^2}\le \frac{1}{x+\pi}$$

whereupon applying the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \pi}\frac{\sin(x-\pi)}{x^2-\pi^2}=\frac{1}{2\pi}}$$

  • $\begingroup$ Thanks for a new approach to this problem! $\endgroup$
    – Quant
    Apr 15 '16 at 21:10
  • $\begingroup$ @Quant You're welcome! My pleasure. I really just wanted to give you the best answer I could. -Mark $\endgroup$
    – Mark Viola
    Apr 15 '16 at 22:02

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.