Question

Evaluate:

$\rm \displaystyle \lim_{x\to \pi}\ \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin^{2} x}$


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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \: \displaystyle \lim_{x\to \pi}\ \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin^{2} x}$

If we substitute directly the value of x, we get

$\rm \: = \: \: \ \dfrac{\sqrt{1-\cos \pi}-\sqrt{2}}{\sin^{2} \pi}$

$\rm \: = \: \: \ \dfrac{\sqrt{1- ( - 1)}-\sqrt{2}}{0}$

$\rm \: = \: \: \ \dfrac{\sqrt{2}-\sqrt{2}}{0}$

$\rm \: = \: \: \ \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle \lim_{x\to \pi}\ \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin^{2} x}$

On rationalizing the denominator, we get

$\rm \: = \: \displaystyle \lim_{x\to \pi}\ \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin^{2} x} \times \frac{ \sqrt{1 - cosx} + \sqrt{2} }{ \sqrt{1 - cosx} + \sqrt{2} }$

$\rm \: = \: \displaystyle \lim_{x\to \pi}\ \frac{ {( \sqrt{1 - cosx} )}^{2} - ( { \sqrt{2} )}^{2} }{\sin^{2} x( \sqrt{1 - cosx} + \sqrt{2} )}$

$\rm \: = \: \displaystyle \lim_{x\to \pi} \: \frac{1}{\sqrt{1 - cosx} + \sqrt{2} } \times \displaystyle \lim_{x\to \pi}\ \frac{1 - cosx - 2 }{\sin^{2} x}$

$\rm \: = \: \: \frac{1}{\sqrt{1 - cos\pi} + \sqrt{2} } \times \displaystyle \lim_{x\to \pi}\ \frac{ - cosx - 1}{1 - \cos^{2} x}$

$\rm \: = \: \: \frac{1}{\sqrt{1 - ( - 1)} + \sqrt{2} } \times \displaystyle \lim_{x\to \pi}\ \frac{ - (cosx + 1)}{(1 - cosx)(1 + cosx}$

$\rm \: = \: \: \frac{1}{\sqrt{2} + \sqrt{2} } \times \displaystyle \lim_{x\to \pi}\ \frac{ - 1}{(1 - cosx)}$

$\rm \: = \: \: \frac{1}{2\sqrt{2} } \times \ \frac{ - 1}{(1 - cos\pi)}$

$\rm \: = \: \: \frac{1}{2\sqrt{2} } \times \ \frac{ - 1}{1 - ( - 1)}$

$\rm \: = \: \: \frac{1}{2\sqrt{2} } \times \ \frac{ - 1}{2}$

$\rm \: = \: \: - \: \frac{1}{4\sqrt{2} }$

Hence,

$\rm\implies \: \: \boxed{\sf{ \:\: \displaystyle \lim_{x\to \pi}\ \frac{\sqrt{1-\cos x}-\sqrt{2}}{\sin^{2} x} = \: - \: \frac{1}{4 \sqrt{2} } \: \: }}$

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Formulae Used :-

$\boxed{\sf{ \:cos\pi \: = \: - \: 1 \: }}$

$\boxed{\sf{ \:sin\pi \: = \: \: 0 \: }}$

$\boxed{\sf{ \: \: {sin}^{2}x + {cos}^{2}x = 1 \: \: }}$

$\boxed{\sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2} \: \: }}$

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Additional Information

$\boxed{\sf{ \:\displaystyle \lim_{x\to 0} \: \frac{sinx}{x} = 1 \: }}$

$\boxed{\sf{ \:\displaystyle \lim_{x\to 0} \: \frac{tanx}{x} = 1 \: }}$

$\boxed{\sf{ \:\displaystyle \lim_{x\to 0} \: \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{\sf{ \:\displaystyle \lim_{x\to 0} \: \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{\sf{ \:\displaystyle \lim_{x\to 0} \: \frac{ {a}^{x} - 1}{x} = loga \: }}$

Answer 2

Answer:

ello

gud mrng

hru

(≡^∇^≡)