Question

lim x tends to π/2 tan5x/tanx

Answer 1

lim (x ->0) pi/2 tan5x/tanx = 
lim (x ->0) pi/2 (sin5x/sinx)*(cosx/cos5x)
lim (x ->0) pi/2 (sin5x/sinx) | h'ospital rule |
lim (x ->0) pi/2 (5*cos5x/cosx) = 5π/2

Answer 2

The limit is equal to $\dfrac{1}{5}$ or 0.2.

Step-by-step explanation:

Given:

The limit is given as:

$\lim_{x \to \frac{\pi}{2}} \frac{\tan 5x}{\tan x}$

We know,

$\tan x=\frac{\sin x}{\cos x}$

∴ $\tan 5x=\frac{\sin 5x}{\cos 5x}$

$\lim_{x \to \frac{\pi}{2}} (\frac{\sin 5x}{\cos 5x}\div \frac{\sin x}{\cos x})\\\\\lim_{x \to \frac{\pi}{2}} (\frac{\sin 5x}{\cos 5x}\times \frac{\cos x}{\sin x})\\\\\lim_{x \to \frac{\pi}{2}} (\frac{\sin 5x}{\sin x}\times \frac{\cos x}{\cos 5x})\\\\\lim_{x \to \frac{\pi}{2}} \frac{\sin 5x}{\sin x}\times \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\cos 5x}\\\\\frac{\sin (\frac{5\pi}{2})}{\sin \frac{\pi}{2}}\times \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\cos 5x}$

$1\times \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\cos 5x}\\\\\textrm{Applying L'Hospital Rule, we get}\\\\=\lim_{x \to \frac{\pi}{2}} \frac{-\sin x}{-5\sin 5x}\\\\=\frac{-\sin(\frac{\pi}{2})}{-5\sin{(\frac{5\pi}{2})}}\\\\=\frac{1}{5}=0.2$

Therefore, the limit is equal to $\frac{1}{5}$ or 0.2.