Question

Lim x tends to O xtanx/1-cosx
Plzzzzz give the answer its urgent and plzz solve it step wise

Answer 1

I hope it helps uh..

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Answer 2

The value of $\lim_{x \to 0} \dfrac{x\tan x}{1-\cos x}$ is equal to 2.

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{x\tan x}{1-\cos x}$

To find, the value of $\lim_{x \to 0} \dfrac{x\tan x}{1-\cos x}$ = ?

∴ $\lim_{x \to 0} \dfrac{x\tan x}{1-\cos x}$

$=\lim_{x \to 0} \dfrac{x\sin x}{\cos x(1-\cos x)}$

Multiply numerator and denominator by $\sin x$, we get

$=\lim_{x \to 0} \dfrac{x\sin^2 x}{\sin x\cos x(1-\cos x)}$

Using the trigonometric identity,

$\sin^2 A+\cos^2 A=1$

⇒ $\sin^2 A=1-\cos^2 A$

$=\lim_{x \to 0} \dfrac{x(1-\cos^2 x)}{\sin x\cos x(1-\cos x)}$

$=\lim_{x \to 0} \dfrac{x(1+\cos x)(1-\cos x)}{\sin x\cos x(1-\cos x)}$

$=\lim_{x \to 0} \dfrac{x(1+\cos x)}{\sin x\cos x}$

$=\lim_{x \to 0} \dfrac{(1+\cos x)}{\dfrac{\sin x}{x} \cos x}$

$=\lim_{x \to 0} \dfrac{(1+\cos x)}{1. \cos x}$

= $\dfrac{(1+\cos 0)}{1. \cos 0}$

= $\dfrac{1+1}{1}$

= 2

∴ The value of $\lim_{x \to 0} \dfrac{x\tan x}{1-\cos x}$ = 2

Thus, the value of $\lim_{x \to 0} \dfrac{x\tan x}{1-\cos x}$ is equal to 2.