Question

Prove that,

$\displaystyle \mathrm{\lim_{x\to 0}\frac{tanx-x}{x^{2}\:tanx}=\frac{1}{3}}$​

Answer 1

hi mate ☺️. use l hospital rule . Then solve .

Step-by-step explanation:see the attachment below .mark answer as brainlist and follow me.

Answer 2

Answer: 1/3 (Using L' Hospital Rule)

Step-by-step explanation:

Given,

$\lim_{x\to0}\frac{\tan x-x}{x^2\tan x}$

Putting limits, gives an indeterminate value, So,

Applying L' Hospital's Rule,

$=\lim_{x\to0} \frac{\frac{d(\tan x -x)}{dx}}{\frac{d(x^2\tan x)}{dx}}\\\;\\=\lim_{x\to0}\frac{\sec^2x-1}{\tan x.2x+x^2\sec^2x}\\\;\\=\lim_{x\to0}\frac{\tan^2x}{2x\tan x+x^2\sec^2x}\;\;\;\;\;\;\because\tan^2x=\sec^2x-1}\\\;\\\text{Dividing Numerator and Denominator by}\;x^2\\\;\\=\lim_{x\to0}\frac{\frac{\tan^2x}{x^2}}{2.\frac{\tan x}{x}+\sec^2x}\\\;\\=\lim_{x\to0}\frac{\tan x}{x}.\frac{\tan x}{x}.\lim_{x\to0}\frac{1}{2.\frac{\tan x}{x}+\sec^2x}\\\;\\=1\times1.\frac{1}{2+1}\\\;\\=\frac{1}{3}$

$\textbf{Hence Proved}$