Finding $lim_{x \to \pi/8 } \frac {\cot 4x - \cos 4x}{(\pi-8x)^3} $

Question

I have been trying to solve the above limits by reducing the trigonometric expression by substituting for $\cot x $ as $\frac {\cos x}{\sin x} $, But not able to go further from there , any help how to solve this ?

Answer 1

$\lim_{x \to \pi/8 } \frac{\cot 4x – \cos 4x}{(\pi-8x)^3}$

$\lim_{x \to \pi/8 } \frac{\frac {\cos 4x}{\sin 4x} – \cos 4x}{(\pi-8x)^3}$

Put $\pi-8x=h$, therefore, $x=\frac{\pi-h}{8}$

And as $x\to\pi/8$, $h\to0$,

$lim_{h \to 0}[\frac{\cos4(\pi-h)/8}{\sin4(\pi-h)/8}-\cos4(\pi-h)/8]/h^3$

$=lim_{h \to 0}\cos(\pi/2-h/2)[\frac{1}{\sin(\pi/2-h/2)}-1]/h^3$

$=lim_{h \to 0}\sin h/2[\frac{1}{\cos h/2}-1]/h^3$

Putting $h/2=t$, therefore, as $h\to 0$, $t\to 0$

Substituting $h/2$ as $t$ in the above expression gives,

$=lim_{t \to 0}[\sin t.lim_{t \to 0} (\frac{1-\cos t}{\cos t})]/8t^3$

$=\frac{1}{8}.lim_{t \to 0} \frac{\sin t}{t}.lim_{t \to 0}\frac{1}{\cos t}.lim_{t \to 0}\frac{1-cost}{t^2}$

$=\frac{1}{8}.lim_{t \to 0}\frac{1-\cos^2t}{t^2.(1+\cos t)}$

$=\frac{1}{8}.1.1.lim_{t \to 0}\frac{1-\cos t}{t^2(1+\cos t)}$

$=\frac{1}{8}.lim_{t \to 0}\frac{\sin^2t}{t^2}.lim_{t \to 0}\frac{1}{1+\cos t}$

$=\frac{1}{8}.1^2.\frac{1}{1+1} = \frac{1}{8}.\frac{1}{2}=\frac{1}{16}$

Answer 2

Write $$lim_{x\rightarrow \pi/8}\frac{cos4x(1-sin4x)}{sin4x(\pi-8x)^3}$$

$$=lim_{x\rightarrow \pi/8}\frac{cos4x-\frac{sin8x}{2}}{(\pi-8x)^3}$$

$$=lim_{x\rightarrow \pi/8}\frac{2cos4x-sin8x}{2(\pi-8x)^3}$$

$$=lim_{x\rightarrow \pi/8}\frac{sin4x+cos8x}{6(\pi-8x)^2}$$...Using LH-rule

$$=lim_{x\rightarrow \pi/8}\frac{cos4x-2sin8x}{24(8x-\pi)}$$......Using LH-rule

$$=lim_{x\rightarrow \pi/8}\frac{-4sin4x-16cos8x}{24(8)}=\frac{1}{16}$$......Using LH-rule