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

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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

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

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

Answer 4

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

Answer 5

\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}