Find the limit $\lim_{x\to0}\frac{\tan6x)}{\sin3x}$ without using L'hopital

Question

I'm trying to compute the following limit: $$\lim_{x\to0}\frac{\tan6x}{\sin3x}$$

I really have no idea how to start it. I tried rewriting $\tan6x$ in terms of $\sin6x$ and $\cos6x$ but wasn't able to simplify the expression. How do I go about this?

Answer 1

When $\sin(3x)\ne0$, $$ \begin{align} \frac{\tan(6x)}{\sin(3x)} &=\frac{\sin(6x)}{\cos(6x)\sin(3x)}\\ &=\frac{2\sin(3x)\cos(3x)}{\cos(6x)\sin(3x)}\\ &=2\frac{\cos(3x)}{\cos(6x)} \end{align} $$

Answer 2

By standard limit $\frac{\sin t}{t}\to 1$ as $t\to 0$, we have

$$\frac{\tan(6x)}{\sin(3x)}=\frac{\tan(6x)}{6x}\frac{3x}{\sin(3x)}\frac{6x}{3x}\to 1\cdot 1\cdot 2=2$$

Answer 3

With $\tan6x=\dfrac{2\tan3x}{1-\tan^23x}=\dfrac{2\sin3x}{\cos3x(1-\tan^23x)}$ then $$\lim_{x\to0}\dfrac{\tan6x}{\sin3x}=\lim_{x\to0}\dfrac{2}{\cos3x(1-\tan^23x)}=2$$

Answer 4

For fun:

Note: $\tan 6x = \dfrac{2\tan 3x}{1-\tan^2 3x}$.

$\dfrac{1}{\cos 3 x}\dfrac {\tan 6x}{\tan 3x}=$

$\dfrac{1}{ \cos 3x} \dfrac{2}{1-\tan^2 3x}.$

The limit is?

Answer 5

$$ \lim _{x\to 0} \frac {\tan 6x}{\sin 3x} \\= \lim _{x\to 0} \frac {\sin 6x}{\sin 3x\cos 6x}\\=\lim _{x\to 0} \frac {1} {\cos 6x} \lim _{x\to 0} \frac {\sin 6x}{\sin3x}\\ =\lim _{x\to 0} \frac {2\sin 3x \cos 3x}{\sin3x}= \lim _{x\to 0} {2 \cos 3x} =2$$