Limit of $\frac{\sin 2x}{x}$ when $x \to 0$
By Gabriel Cooper •
My teacher solved this:
$$\lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} \frac{2\sin 2x}{2x} = \lim_{x \to 0} \frac{\sin x}{x} = 1$$
I have problem with this part:
$$\lim_{x \to 0} \frac{2\sin 2x}{2x} = \lim_{x \to 0} \frac{\sin x}{x}$$
which is equal to
$$\frac{2 \sin 2x}{2x} = \frac{\sin x}{x}$$
Is there a formula for this?
$\endgroup$ 11 Answer
$\begingroup$$$ \lim_{x \to 0} \frac{ \sin (2x)}{x} = \lim_{x \to 0} \frac{ \sin (2x)}{x} \frac{2}{2} = \lim_{2x \to 0} \frac{ \sin (2x)}{2x} \times 2 = 1 \times 2 = 2$$
If $x$ tends to $0$, then $2x$ also tends to zero.
And the problem follows by using the formula:
$$ \lim_{t \to 0 } \frac{\sin (t)}{t} = 1 $$
In general $$ \lim_{t \to 0} \frac{ \sin (nt) }{t} = n $$
$\endgroup$ 3
