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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?

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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 $$\lim_{x \to \frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)}{(x-\frac{\pi}{2})^3}=\\ \lim_{a\to 0} \frac{\sin(-a)}{(a)^3}\\= \lim_{a\to 0} \frac{-\sin(a)}{(a)^3}\\= \lim_{a\to 0} \frac{-1}{(a)^2}\to -\infty$$

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

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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.

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  • $\begingroup$ 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? $\endgroup$ – skrrrt Jul 12 '20 at 5:39
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In such cases it is often quite practical to shift the limit to $0$.

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

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

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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:

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

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