Question

lim x tends to 0 2sinx- sin 2x ÷ x^3


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Answer 1

I hope this helps you.

Answer 2

The value of $\lim_{x \to 0} \dfrac{2\sin x-\sin 2x}{x^{3} }$is 1.

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{2\sin x-\sin 2x}{x^{3} }$

To find, the value of $\lim_{x \to 0} \dfrac{2\sin x-\sin 2x}{x^{3}}$= ?

=$\lim_{x \to 0} \dfrac{2\sin x-2\sin x\cos x}{x^{3} }$ [$\dfrac{0}{0}$ form]

[ ∵ $\sin 2A=2\sin A\cos A$]

=$\lim_{x \to 0} \dfrac{2\sin x(1-\cos x)}{x^{3} }$

=$2\lim_{x \to 0} \dfrac{\sin x}{x} \dfrac{(1-\cos x)}{x^{2} }$

=$2\lim_{x \to 0} 1. \dfrac{(2\sin ^{2}\frac{x}{2} )}{x^{2} }$

=$4\lim_{x \to 0} 1. \dfrac{(\sin ^{2}\frac{x}{2} )}{(\dfrac{x}{2} )^{2} }\times \frac{1}{4}$

= 4 × $\dfrac{1}{4}$

= 1

Hence, the value of $\lim_{x \to 0} \dfrac{2\sin x-\sin 2x}{x^{3} }$is 1.