Find out the answer to "Solve:limx→0x3tan3(2x)"

oskrnavih92j

Answered question

2022-03-01

Solve:

lim_{x \to 0} \frac{x^{3}}{\tan^{3} (2 x)}

Answer & Explanation

alagmamGynccip

Beginner2022-03-02Added 4 answers

I'll assume that you know that
$lim_{x \to 0} \frac{\sin (x)}{x} = 1$
by knowing this you can show by doing a simple substitution $u = a x$ that
$lim_{x \to 0} \frac{\sin a x}{b x} = \frac{a}{b}$
this will also give you
$lim_{x \to 0} \frac{b x}{\sin (a x)} = \frac{1}{lim_{x \to 0} \frac{\sin (a x)}{b x}} = \frac{b}{a}$
knowning this and since $\tan (x) = \frac{\sin (x)}{\cos (x)}$ you can show that
$lim_{x \to 0} \frac{\tan (x)}{x} = 1$
and
$lim_{x \to 0} \frac{\tan (a x)}{b x} = \frac{a}{b}$
which implies that
$lim_{x \to 0} \frac{b x}{\tan (a x)} = \frac{b}{a}$
that's all you need to solve your question and similar questions since
$lim_{x \to 0} \frac{x^{3}}{\tan^{3} (2 x)} = lim_{x \to 0} \frac{x}{\tan (2 x)} \times lim_{x \to 0} \frac{x}{\tan (2 x)} \times lim_{x \to 0} \frac{x}{\tan (2 x)}$
$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$
$= \frac{1}{8}$

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