# How do you find the limit of (sin2x)/x as x approaches 0?

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#### Explanation

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#### Explanation:

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90
Apr 21, 2017

In the Explanation

#### Explanation:

okay, we have

${\lim}_{x \to 0}$$\frac{\sin 2 x}{x}$

now, if we were to operate it directly, we would get $\frac{0}{0}$ as answer which is indefinite, so we need to operate it more before applying the limiting value

So,

${\lim}_{x \to 0}$$\frac{\sin 2 x}{x}$

$= {\lim}_{x \to 0}$$\frac{2 \sin x \cos x}{x}$ (because $\sin 2 x = 2 \sin x \cos x$)

rearrange it as:

$= {\lim}_{x \to 0}$$\sin \frac{x}{x} \left(2 \cos x\right)$

$= 1 \times 2 \cos \left(0\right)$ (since${\lim}_{x \to 0}$$\sin \frac{x}{x} = 1$)

$= 2 \times 1$ (as $\cos 0 = 1$)

$= 2$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

32
Oct 20, 2015

$2$
Remember : ${\lim}_{x \to 0} \frac{\sin k x}{x} = k$

#### Explanation:

PROOF:

${\lim}_{x \to 0} \frac{\sin k x}{x} = {\lim}_{x \to 0} \left(k \cdot \frac{\sin k x}{k x}\right)$

$= k \cdot {\lim}_{x \to 0} \frac{\sin k x}{k x}$

As $x \to 0$ : $\implies k x \to 0$

$\implies k \cdot {\lim}_{k x \to 0} \frac{\sin k x}{k x}$

$\implies k \cdot 1 = k$

So here in this case $k = 2$

Therefore answer : $2$

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