Question

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

Answer 1

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Answer 2

Solution :

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Simplification :

(√1-cosx - √2)/sin²x

=[(√1-cosx -√2)(√1-cosx +√2)]/[sin²x(√1-cosx+√2]

= [(√1-cosx)² -(√2)²]/[(1-cos²x)(√1-cosx+√2)]

=[1-cosx-2]/[(1-cosx)(1+cosx)(√1-cosx+√2)]

= [-(cosx+1)][(1-cosx)(1+cosx)(√1-cosx+√2)]

After cancellation, we get

= -1/[(1-cosx)(√1+cosx+√2)]-----( 1 )
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Here ,

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

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

[ from ( 1 ) ]

= $\rm \displaystyle \frac{-1}{(1 - cos\ π)(\sqrt{1+ \cos π}+\sqrt 2}$

= $\rm \displaystyle \frac{-1}{(1+1)(\sqrt 2+\sqrt 2)}$

= -1/(2 × 2√2)

= -1/4√2

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