# Evaluating limit of the function at $\frac{\pi}{2}$

I'm trying to solve this

$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}$$

I have tried using the L'Hôpital's rule

But I'm stuck at

$$\lim_{x \to \frac{\pi}{2}} \frac{-\sin{x}}{3(x-\frac{\pi}{2})^2}$$

Since the above equation is not in the $$\frac{0}{0}$$ , $$\frac{\infty}{\infty}$$ or $$\frac{anything}{\infty}$$ form

Then I tried expanding the $$\cos{x}$$ as taylor series at $$x=\frac{\pi}{2}$$. Which on simplifying I am left with

$$\lim_{x \to \frac{\pi}{2}} \frac{-1}{(x - \frac{\pi}{2})^2} + \frac{1}{6} - \frac{1}{120} (x - \frac{\pi}{2})^2 + \frac{(x - \frac{\pi}{2})^4}{5040} - ...$$

and I'm stuck again. How do I proceed ahead?

## 5 Answers

welcome to MSE

AS a hint

$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3}=\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}$$now take $$x-\frac{\pi}{2}=a$$ when $$x$$ tends to $$\frac{\pi}{2}$$ ,a tends to zero $$\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}=\\ \lim_{a\to 0} \frac{\sin(-a)}{(a)^3}\\= \lim_{a\to 0} \frac{-\sin(a)}{(a)^3}\\= \lim_{a\to 0} \frac{-1}{(a)^2}\to -\infty$$

The limit is $$-\infty$$. Numerator tends to $$-1$$ and denominator tends to $$0$$ through positive values.

Setting $$y=x-\pi/2$$, this is $$\lim_{y\to0}\frac{\cos(y+\pi/2)}{y^3}=\lim_{y\to0}\frac{-\sin y}{y^3}.$$ But $$\frac{-\sin y}{y^3}=-\left(\frac{\sin y}y\right)\left(\frac1{y^2}\right)$$ and the first bracket converges to $$1$$ as $$y\to0$$ and the second bracket diverges to $$\infty$$. So your "limit" is $$-\infty$$ if you regard such things as limits.

• Thank you for the answer. However I'm not able to understand how does the second bracket diverges to infinity. I'm thinking that it would be 1/0 as y ->0 that is not defined. Can you please explain? – skrrrt Jul 12 '20 at 5:39

In such cases it is often quite practical to shift the limit to $$0$$.

Set $$x=t+\frac{\pi}2$$:

$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{x}}{(x-\frac{\pi}{2})^3} = \lim_{t \to 0} \frac{-\sin{t}}{t^3}$$ $$=\lim_{t \to 0}\left(\frac{-\sin t}{t}\cdot\frac 1{t^2}\right)=-\infty$$

Since we have an indeterminate form of type $$(0/0)$$, we can apply the l'Hopital's rule:

$$\color{blue}{\lim_{x \to \frac{\pi}{2}} \frac{\cos{\left(x \right)}}{\left(x - \frac{\pi}{2}\right)^{3}}} = \color{magenta}{\lim_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}\left(\cos{\left(x \right)}\right)}{\frac{d}{dx}\left(\left(x - \frac{\pi}{2}\right)^{3}\right)}}$$

Hence we have:

$$\lim_{x \to \frac{\pi}{2}}\left(- \frac{\sin{\left(x \right)}}{3 \left(x - \frac{\pi}{2}\right)^{2}}\right)=\lim_{x \to \frac{\pi}{2}}\left(- \frac{4 \sin{\left(x \right)}}{3 \left(\pi - 2 x\right)^{2}}\right)$$

After if $$x\to \pi/2$$ we have $$\text{limited function}/0\to +\infty$$ being $$3 \left(\pi - 2 x\right)^{2}\geq0$$, but with minus sign we have:

$$\lim_{x \to \frac{\pi}{2}} \frac{\cos{\left(x \right)}}{\left(x - \frac{\pi}{2}\right)^{3}} = -\infty$$