# Computing $\lim_{h\to 0}\frac{1}{h}\left[\frac{1}{\sin(\frac{\pi}{4}+h)}-\frac{1}{\sin\frac{\pi}{4}}\right]$

I'm currently trying to solve the following exercise:

Compute $$\lim_{h\to 0}\frac{1}{h}\left[\dfrac{1}{\sin(\frac{\pi}{4}+h)}-\dfrac{1}{\sin\frac{\pi}{4}}\right]$$

My approach so far: $$\lim_{h\to 0}\frac{1}{h}\left[\dfrac{1}{\sin(\frac{\pi}{4}+h)}-\dfrac{1}{\sin\frac{\pi}{4}}\right] = \lim_{h\to 0} \frac{1}{h} \left[\dfrac{\sin\frac{\pi}{4}-\sin(\frac{\pi}{4}+h)}{\sin(\frac{\pi}{4}+h)\cdot\sin\frac{\pi}{4}}\right]$$

and now I don't know how to continue or whether I should have chosen a different approach or not.

I have also searched on MSE but didn't find anything similar.

Thank you very much in advance.

• Have you learnt derivatives ?
– user65203
Commented Dec 18, 2019 at 9:54

Without derivative:

$$\lim_{h\to 0} \left(\dfrac{\sin\frac{\pi}{4}-\sin(\frac{\pi}{4}+h)}{h\sin(\frac{\pi}{4}+h)\sin\frac{\pi}{4}}\right)=\frac{\lim_{h\to 0}\frac{ \sin\frac\pi4-\sin\frac\pi4\cos h-\cos\frac\pi4\sin h}h}{\lim_{h\to 0}\sin(\frac{\pi}{4}+h)\sin\frac{\pi}{4}}=-\frac{\cos\tfrac\pi4}{\frac12}\lim_{h\to 0}\frac{\sin h}h=-\sqrt2.$$

Let $$f(x)=\dfrac{1}{\sin x}$$ and $$a=\frac{\pi}{4}$$, then the limit becomes: $$\lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}=f'(a)$$ As $$f'(x)=-\dfrac{\cos x}{\sin^2x}$$, therefore the answer is $$f'(\dfrac{\pi}{4})=-\sqrt{2}$$

If your teacher doesn't accept you to use derivatives, you can use the another way.

By using $$\sin A-\sin B=2\cos\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$$, I continue your process: $$\lim_{h\to 0}\dfrac{1}{h}\left[\dfrac{2\cos(\frac{\pi}{4}+\frac{h}{2})\sin(\frac{-h}{2})}{\sin^2\frac{\pi}{4}}\right]=-\sqrt{2}\lim_{h\to 0}\dfrac{\sin\frac{h}{2}}{\frac{h}{2}}=-\sqrt{2}$$

The limit is the derivative of $$\frac 1 {\sin x}$$ at $$\pi /4$$. So it is $$-\frac 1 {\sin^{2}(\pi /4)} \cos (\pi /4) =-\sqrt 2$$. [$$\sin (\pi/4)=\cos (\pi/4)=\frac 1 {\sqrt 2}]$$.

Use L'Hospital's method to find the limit $$L=\lim_{h \rightarrow 0}\frac{-1}{h}~~\frac{\sin(\pi/4+h)-\sin(\pi/4)}{\sin(\pi/4+h)\sin(\pi/4)}$$ $$=\lim_{h \rightarrow 0} \frac{-1}{h} \frac {2 \cos (\pi/4+h/2) \sin (h/2)}{\sin(\pi/4+h) \sin(\pi/4)}= \lim_{h \rightarrow 0}-\frac{\frac{\sin(h/2)}{h/2} \cos(\pi/4)}{\sin^2(\pi/4)}=-\sqrt{2}.$$