Question

lim 1-cos2x÷1-cos4x
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Answer 1

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Answer 2

SOLUTION

COMPLETE QUESTION

$\displaystyle \sf\lim_{x \to 0} \: \frac{1 - \cos 2x}{1 - \cos 4x}$

EVALUATION

Here the given expression is

$\displaystyle \sf\lim_{x \to 0} \: \frac{1 - \cos 2x}{1 - \cos 4x}$

We solve it using L'HOSPITAL RULE

$\displaystyle \sf\lim_{x \to 0} \: \frac{1 - \cos 2x}{1 - \cos 4x} \: \: \bigg( \frac{0}{0} \: \: form \bigg)$

$\displaystyle \sf = \lim_{x \to 0} \: \frac{ + 2 \sin 2x}{ + 4 \sin 4x} \: \: \: \: \bigg( \frac{0}{0} \: \: form \bigg)$

$\displaystyle \sf = \lim_{x \to 0} \: \frac{ + 4 \cos 2x}{ + 16 \cos 4x}$

$\displaystyle \sf = \frac{ 4 \times 1}{ 16 \times 1}$

$\displaystyle \sf = \frac{ 4 }{ 16}$

$\bf = \dfrac{1}{4}$

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