Question

Evaluate:
$\rm \displaystyle \lim_{x\to \pi/2}\ \frac{\sqrt{3+\sin x}-2}{\cos^{2} x}$

Answer 1

Solution :

_______________________

Simplification :

( √3+sinx - 2 )/(cos²x)

=[(√3+sinx-2)(√3+sinx+2)]/[(1-sin²x)(√3+sinx+4)

= [(√3+sinx)² - 2²]/[(1+sinx)(1-sinx)(√3+sinx+4)]

After cancellation, we get

= (-1)/[(1+sinx)(√3+sinx + 1 )] ----( 1 )
________________________

Here ,

$\rm \displaystyle \lim_{x\to \pi/2}\ \frac{\sqrt{3+\sin x}-2}{\cos^{2} x}$

= $\rm \displaystyle \lim_{x\to \pi/2}\ \frac{-1}{(1+sinx)\sqrt{3+\sin x}+2}$

[from ( 1 ) ]

= (-1)/[( 1 + 1 )( √(3+1) +2 ]

= -1/[ 2 ×( 2 + 2 )]

= -1/8

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