Evaluating $ \lim_{x\to0} \frac{\tan(2x)}{\sin x}$ $$\lim_{x\to0} \frac{\tan(2x)}{\sin x}$$
How would I evaluate that? I was thinking changing the tan to sin/cos, but when I tried that, it did not work.
 A: HINT : $$\frac{\tan (2x)}{\sin x}=\frac{\tan(2x)}{\color{red}{2x}}\cdot \frac{\color{red}{x}}{\sin x}\cdot \color{red}{2}.$$
A: Use $$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin(x)\cos(x)}{\cos(x)^2-\sin(x)^2}$$
Source: http://mathworld.wolfram.com/Double-AngleFormulas.html
A: $$\lim_{x\to0}\frac{\tan (2x)}{\sin x}=\lim_{x\to0}\frac{\sin (2x)}{\cos(2x)\sin x}=\lim_{x\to0}\frac{2\sin x\cos x}{\cos(2x)\sin x}=\lim_{x\to0}\frac{2\cos x}{\cos(2x)}=2$$
A: Hint:
$$
\frac{\tan (2x)}{\sin x} = 
\frac{\sin(2x)}{\cos(2x)} \cdot\frac{1}{\sin x} = 
2 \cdot\frac{\sin(2x)}{(2x)} \cdot \frac{x}{\sin x} \cdot \frac 1{\cos(2x)}
$$
A: I think something's wrong with our pedagogy when students have to ask about things like this (but of course a lot is already known to be wrong with our pedagogy, in ways that make this predictable.
If one knows that $\left.\dfrac{d}{dx}\tan(2x)\right|_{x=0}=2$, then one knows that the function is locally like $x\mapsto 2x$ near $x=0$.  Similarly $\sin x$ behaves like $1\cdot x$ near $x=0$.  And since $\lim\limits_{x\to0}$ is about local behavior near $x=0$, so $\tan(2x)/\sin x$ is like $2x/x$; hence the limit is $2$.
Not only do students come out of calculus courses with good grades without having been taught these ideas, but they come out never having heard that it's not all about memorizing meaningless algorithms for the purpose of proving one's skills at obeying complicated instructions.
A: You can use Maclaurin series;
The series representation centered at $0$ of $\frac {\tan(2x)}{sinx}$ is $$2+3x^2+\frac {19x^4}{4}+\frac {307x^6}{40}+\mathcal{O}(x^7)$$
All the terms become $0$ but the $2$.
I don't think the evaluation of the serirs representation would be easier than other methods to evaluate the limit though.
A: You can use the identity 
$$\tan(2x) = \frac{2\tan x}{1-\tan^2x}$$
A: Since $\displaystyle\tan(2x)=\frac{2\tan x}{1-\tan^2x}$
Set $t=\frac x2$
Using tangent half-angle formula we get;
$\tan x=\frac{2t}{1-t^2}$, $\quad$$\sin x=\frac{2t}{1+t^2}$
Hence $\displaystyle\lim\limits_{x\to 0}\frac{\tan(2x)}{\sin(x)}=\displaystyle\lim\limits_{x\to 0}\frac{2\tan x}{1-\tan^2x}\cdot\frac{1}{\sin x}$
$\displaystyle=\lim\limits_{t\to 0}\frac{2\big(\frac{2t}{1-t^2}\big)}{1-\big(\frac{2t}{1-t^2}\big)^2}\cdot\frac{1+t^2}{2t}=2\lim\limits_{t\to 0}\frac{1-t^4}{1+t^4-6t^2}=2\cdot 1=2$
