Question

evaluate limx-0 sin4x/sin2x

Answer 1

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Answer 2

$\lim_{x \to \ 0} \dfrac{\sin 4x}{\sin 2x} = 2$

Step-by-step explanation:

We have to calculate the value of the limit.

$\lim_{x \to \ 0} \dfrac{\sin 4x}{\sin 2x}$

= $\lim_{x \to \ 0} \dfrac{\dfrac{\sin 4x}{4x}}{\dfrac{\sin 2x}{2x}} \times 2$

{Since $x \to \ 0$, so $x\neq 0$ and hence, we can divide by 4x or 2x}

= $2 \times \dfrac{ \lim_{4x \to \ 0} \frac{\sin 4x}{4x}}{ \lim_{2x \to \ 0} \frac{\sin 2x}{2x}}$

{Since we know the formula of limit as $\lim_{x \to \ 0} \dfrac{\sin x}{x} = 1$ }

= $2 \times \dfrac{1}{1}$ { As $x \to \ 0$ so, $4x \to \ 0$ and also $2x \to \ 0$ }

= 2 (Answer)