The value of limx→π/2 tan2⁡x2sin2⁡x+3sin⁡x+4−sin2⁡x+6sin⁡x+2 is equal to

A
$\frac{1}{10}$
B
$\frac{1}{11}$
C
$\frac{1}{12}$
D
$\frac{1}{8}$

Solution:

We have,

$\begin{array}{l} lim_{x \to π / 2} \tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}) \\ = lim_{x \to π / 2} \frac{\tan^{2} x (2 \sin^{2} x + 3 \sin x + 4 - \sin^{2} x - 6 \sin x - 2)}{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2}} \end{array}$

$\begin{array}{l} = lim_{x \to π / 2} \frac{\tan^{2} x (\sin^{2} x - 3 \sin x + 2)}{\sqrt{2 \sin^{2} x + 3 \sin x + 4 + \sqrt{\sin^{2} x + 6 \sin x + 2}}} \\ = lim_{x \to π / 2} \frac{\sin^{2} x (\sin x - 1) (\sin x - 2)}{(1 - \sin^{2} x) \{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2})} \\ = lim_{x \to π / 2} \frac{- \sin^{2} x (\sin x - 2)}{(1 + \sin x) \{\sqrt{2 \sin^{2} x + 3 \sin x + 4} + \sqrt{\sin^{2} x + 6 \sin x + 2}} \\ = \frac{1}{2 (\sqrt{9} + \sqrt{9})} = \frac{1}{12} \end{array}$

The value of $lim_{x \to π / 2} \tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2})$ is equal to

Solution:

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