Question

Lim xtends to 0 xtanx/1-cosx

Answer 1

let me know whether it is right or not...

Answer 2

$\fontsize{18}{10}{\textup{\textbf{The value of the limit is 2.}}}$

Step-by-step explanation:

Here, we will using L'Hospital's rule.

The given limit is

$L=\lim_{x\rightarrow 0}\dfrac{x\tan x}{1-\cos x}.$

When x = 0, then x tan x = 1-\cos x = 0, so the given fraction is of the form 0/0.

Applying L'Hospital's rule, we have

$L\\\\\\=\lim_{x\rightarrow 0}\dfrac{x\tan x}{1-\cos x}\\\\\\=\lim_{x\rightarrow0}\dfrac{x\sec^2x+\tan x}{\sin x}~~~~~[\frac{0}{0}]\\\\\\=\lim_{x\rightarrow 0}\dfrac{\sec^2x+2x\sec^2x\tan x+\sec^2x}{\cos x}\\\\\\=\lim_{x\rightarrow 0}\dfrac{2\sec^2x+2x\sec^2x\tan x}{\cos x}\\\\\\=\dfrac{2\times1+0}{1}\\\\=2.$

Thus, the required value of the limit is 2.

#Learn more

Question : Limit x tends to 0 sinax/tanbx.

Link : https://brainly.in/question/1820978.