# Evaluate $\lim _{x\to \pi }\left(\frac{\sin x}{\pi ^2-x^2}\right)$ without L'Hopital or Taylor series

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 $x \to \pi$.

• Oh, thank you! I was so stupid not to have noticed it! – Quant Apr 15 '16 at 19:04
• Nice. Its a learning experience... – DeepSea Apr 15 '16 at 19:04
• 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? – IgNite Apr 15 '16 at 19:43
• @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 – 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

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

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