How is $\lim_{x \to 0} \frac{\sin x}{x} = 1$ used to compute $\lim_{x \to 0} \frac{\sin 2x}{x}$? I know that I can substitute $2x$ for $x$ in 
$$\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$$
to obtain the following equality:
$$\displaystyle\lim_{2x \to 0} \frac{\sin 2x}{2x} = 1$$
But the slide below seems to suggest that 
$$\displaystyle\lim_{2x \to 0} \frac{\sin 2x}{2x} = 1 \Longleftrightarrow  \displaystyle\lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \tag{1}$$
(note that $2x \to 0$ becomes $x \to 0$). 
What's the justification for this?
Furthermore, what's the justification for 
$$ \lim_{x \to 0} \left( 2 \frac{\sin(2x)}{2x}\right) = 2 \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} \tag{2}$$ ?

 A: As $2x\to0$, $x\to0$ as well. This is because of what the "limit" really means -- it asks you to evaluate the expression as $2x$ gets smaller and smaller. But if you think about it, this is really the same thing as $x$ getting smaller and smaller. Thus the two expressions will be the same.
Pedantic note, obviously you also want to make sure it's true if $x\to0^-$ as well.
A: Another way to do it is to expand $\sin (2x)$ as $ 2 \sin (x) \cos(x)$. 
The limit then becomes: 
$$\lim_{x\to 0} \frac{2 \sin (x) \cos(x)}{x}$$
$$=2$$
A: Basically, what $\lim\limits_{x \to 0} \frac{\sin(x)}{x} = 1$ says is that the ratio of the sine of a value to its own value approaches $1$, as the value approaches $0$.
So, if you look at $\lim_\limits{x \to 0} \frac{\sin(2x)}{2x}$, you can clearly see that as $x$ tends to $0$, the product of any factor ($2$ in this case) and $x$ must also tend to $0$. So this limit also involves the ratio of the sine of a value to its own value, as the value approaches $0$. Hence, this limit is equivalent to $\lim\limits_{2x \to 0} \frac{\sin(2x)}{2x}$, which in turn is equivalent to $\lim\limits_{x \to 0} \frac{\sin(x)}{x}$.
For your second question, $2$ is a constant factor that is independent of the limit (since it is independent of $x$). Hence, pulling it out the limit is justified.
