"Find:limx→01−cos⁡4xxtan⁡6x" solved and explained

George Michael

Answered question

2022-04-07

Find:

lim_{x \to 0} \frac{1 - \cos 4 x}{x \tan 6 x}

Answer & Explanation

wyjadaczeqa8

Beginner2022-04-08Added 14 answers

By using $lim_{x \to 0} \frac{\sin^{2} 4 x}{16 x^{2}} = 1$ and $lim_{x \to 0} \frac{6 x}{\tan 6 x} = 1$ we can find that you have the limit if your expression approximation like $\frac{(\frac{1 - (1 - (4 x)^{2}}{2})}{x \times 6 x}$ when x is close enought to 0. And we have that limit is equivalent to $\frac{8 x^{2}}{6 x^{2}}$ and that is $\frac{4}{3}$
From this, we can work it out and simplify to get
$\frac{1 - \cos 4 x}{x \tan 6 x}$
$= \frac{2 \sin^{2} 2 x \times \cos 6 x}{x \sin 6 x}$
$= 2 \times {(2 \times \frac{\sin 2 x}{2 x})}^{2} \times \frac{1}{\frac{\sin 6 x}{6 x} \times 6} \times \cos 6 x$
$= \frac{2 \sin^{2} (2 x) \cot (6 x)}{x}$
Since we know that $\frac{2 \sin^{2} (2 x) \cot (6 x)}{x}$ is the simplification of the trigonometric limit, we must take the limit of this result to find the answer to the once before limit.
$lim_{x \to 0} \frac{2 \sin^{2} (2 x) \cot (6 x)}{x} = \frac{4}{3}$

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