Question

Evaluate:
$\rm \displaystyle \lim_{x\to 0}\ \frac{1-\cos 2x}{\sin^{2} 2x}$

Answer 1

Solution :

$\rm \displaystyle \lim_{x\to 0}\ \frac{1-\cos 2x}{\sin^{2} 2x}$

= $\rm \displaystyle \lim_{x\to 0}\ \frac{1-(1-2\sin^{2} x)}{4\sin^{2} xcos^{2} x}$

= $\rm \displaystyle \lim_{x\to 0}\ \frac{2\sin^{2} x}{4\sin^{2} xcos^{2} x}$

= $\rm \displaystyle \lim_{x\to 0}\ \frac{2}{4cos^{2} x}$

= $\rm \displaystyle \frac{2}{4cos^{2} 0}$

= 2/( 4 × 1 )

= 2/4

= 1/2

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