# Evaluating $\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$ without L'Hopital

I need to calculate the following limit (without using L'Hopital - I haven't gotten to derivatives yet):

$$\lim_{x\rightarrow\pi}\frac{\sin x}{x^2-\pi ^2}$$

We have $\sin$ function in the numerator so it looks like we should somehow make this similair to $\lim_{x\rightarrow 0} \frac{\sin x}{x}$. When choosing $t=x^2-\pi ^2$ we get $\lim_{t\rightarrow 0} \frac{\sin \sqrt{t+\pi ^2}}{t}$ so it's almost there and from there I don't know what to do. How to proceed further? Or maybe I'm doing it the wrong way and it can't be done that way?

• $x^2 - \pi^2 = (x+\pi)(x-\pi)$ – Michael Biro Jan 31 '15 at 13:45

Choosing the substitution $x - \pi = t$, with $t \to 0$, we have $$\lim_{x \to \pi}\frac{\sin x}{x^2 - \pi^2} = \lim_{t \to 0}\frac{\sin(t + \pi)}{t(t + 2\pi)} = \lim_{t \to 0}-\frac{\sin t}{t(t + 2\pi)} = -\frac1{2\pi}$$

Do the substitution $\pi-x=t$, so your limit becomes $$\lim_{t\to0}\frac{\sin(\pi-t)}{-t(2\pi-t)}= \lim_{t\to0}\frac{\sin t}{-t(2\pi-t)}$$ which is elementary.

Consider we have $x^2-\pi^2=(x+\pi)(x-\pi)$ and we know that $\sin(x-\pi)=-\sin(x)$ yielding: $$\frac{-\sin(x-\pi)}{x-\pi}\cdot\frac1{x+\pi}\to -\frac1{2\pi}$$

\begin{align} \frac{\sin(x)}{x^2 - \pi^2} = \frac{x}{x^2(1-\pi^2/x^2)} \prod_{n=1}^{\infty}\left(1 -\frac{x^2}{n^2\pi^2}\right) = \frac{1}{x} \frac{1-x^2/\pi^2}{1-\pi^2/x^2} \prod_{n=2}^{\infty} \left(1 -\frac{x^2}{n^2\pi^2}\right) \end{align} Since no L'Hospital available, we consider of the sign of numerator and denominator as $x \to \pi^{+}$ and $x \to \pi^-$ to obtain \begin{align} \lim_{x\to \pi}\frac{1-x^2/\pi^2}{1-\pi^2/x^2} = -1 \end{align} and hence \begin{align} \lim_{x\to \pi }\frac{\sin(x)}{x^2 - \pi^2} = -\frac{1}{\pi} \prod_{n=2}^{\infty} \left(1 -\frac{1}{n^2}\right) = -\frac{1}{2\pi} \end{align}

\begin{align} =& \lim_{x \to \pi} \left[ \frac{\sin x - \sin \pi}{(x-\pi)(x+\pi)} \right] \\ =& \lim_{x \to \pi} \left[ \frac{\sin x - \sin \pi}{x-\pi} \times \frac{1}{x+\pi} \right] \\\end{align}

Note that the fraction on the left is almost the definition of the derivative for sine. Continuing,

\begin{align} =&\ \cos (x = \pi) \times \lim_{x \to \pi} \left[ \frac{\cos x}{x+\pi} \right] \\ =& -\frac{1}{2\pi} \end{align}

• You're using the derivative. – egreg Jan 31 '15 at 13:48
• Other methods are using equal concept too. $lim_{x\to 0}\frac{sinx}{x}=1$ is not different from derivative for me. They are both connected to each other. Also, using Hopital is not so far from any of these two concept. There is no other method to be so far from Hopital! – Arashium Jan 31 '15 at 13:50
• From OP: “I haven't gotten to derivatives yet”. – egreg Jan 31 '15 at 13:55
• @egreg Point taken. Though it's kind of weird introducing sine without derivatives. – user2345215 Jan 31 '15 at 13:55