Trigonometry - Limits yet again I am having the solution, but don't know why it works:
\begin{align*} \quad \lim_{x \to 0}\frac{\tan(2x)}{x}&=  
  \lim_{x \to 0}\frac{\sin(2x)}{x\cos(2x)}\\
&= \lim_{2x \to 0}\frac{\sin(2x)}{2x} \cdot \lim_{x \to 0}\frac{2}{\cos(2x)}\\ 
&= 1 \cdot 2\\
&=2 \end{align*}
I don't understand why I am allowed to turn \begin{align*} \lim_{x \to 0} \end{align*} into \begin{align*} \lim_{2x \to 0} \end{align*} I am learning this on my own for a entry exam for university and feel a bit lost...
 A: Ok I'm trying to justify the conversion with just logic.Suppose x is negligibly small even 2x will be negligibly small,and for that matter any nx is negligibly small where n is a natural number (unless n tends to infinity),so the conversion you stated is valid.
A: We can prove this works. Consider the $\delta$-$\epsilon$ definition of limit
as found, for example, at
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html
Now suppose you have already shown that 
$$\lim_{2x \to 0}\frac{\sin(2x)}{2x} = 1.$$
Then (by the definition of limit)
you know that for any positive number $\epsilon$, 
there is a number $\delta$ such that
$\left|\frac{\sin(2x)}{2x} - 1\right| < \epsilon$
whenever $0 < 2x < \delta$.
But if $0 < x < \frac\delta2$, then $0 < 2x < \delta$,
so you have just proved the following statement:
For any positive number $\epsilon$,
there is a number $\frac\delta2$ such that
$\left|\frac{\sin(2x)}{2x} - 1\right| < \epsilon$
whenever $0 < x < \frac\delta2$.
But that's just what the definition of $\lim_{x \to 0}\frac{\sin(2x)}{2x} = 1$
says, except that the last paragraph says there is a number $\frac\delta2$
instead of a number $\delta$.  The definition doesn't actually "care"
how you write each number as long as you use the same value wherever the
definition uses the same value, so you have shown that
$$\lim_{x \to 0}\frac{\sin(2x)}{2x} = 1.$$
As noted in comments, you can generalize this to a limit taken as some
continuous, invertible function of $x$ goes to zero, 
where the function $x\mapsto2x$
is just one example of the more general fact.
A: Functions $ \sin x, \tan x $ are straight for small values of $x$, they can be approximated by x as they behave like $y=x$ in the neighborhood of x=0.So,
$$ \dfrac{\sin 2 x}{x} \approx 2,\, \dfrac{\tan 3 x}{\tan 7 x} \approx \frac37  $$
A function like $$ \dfrac{\sin 3.2 x + \tan 6.8 x}{2.5 x} \approx 4.$$  
A: $$\lim_{x \to 0}\frac{\tan(2x)}{x}=  
  \lim_{x \to 0}\frac{\sin(2x)}{x\cos(2x)}$$
You have to make a change of variable. Let's say $y=2x$ or $\displaystyle x= \frac{y}{2}.$
when $x \to 0$ also $y \to 0$.
So we have the following limit: 
$$\lim_{y \to 0}{\frac{\sin y}{\frac{y}{2}\cos y}}$$
$$\lim_{y \to 0}{\frac{2 \sin y}{y\cos y}}.$$
Is it OK?
