How is $\lim_{x \to 0} \frac{\sin x}{x} = 1$ used to compute $\lim_{x \to 0} \frac{\sin 2x}{x}$?

Question

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

$$ \lim_{x \to 0} \left( 2 \frac{\sin(2x)}{2x}\right) = 2 \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} \tag{2}$$ ?

Answer 1

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.

Answer 2

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.

Answer 3

Another way to do it is to expand $\sin (2x)$ as $ 2 \sin (x) \cos(x)$.

The limit then becomes:

$$\lim_{x\to 0} \frac{2 \sin (x) \cos(x)}{x}$$ $$=2$$