calculus - Extended $lim_{x rightarrow 0}{frac{sin(x)}{x}} = 1$ limit law?

So I've learned that $\lim_{x \rightarrow 0}{\frac{\sin(x)}{x}} = 1$ is true and the following picture really helped me get an intuitive feel for why that is

I have been told that this limit is true whenever the argument of sine matches the denominator and they both tend to zero. That is,

$$\lim_{x \rightarrow 0}{\frac{\sin(5x)}{5x}} = 1$$

$$\lim_{x \rightarrow 0}{\frac{\sin(x^2)}{x^2}} = 1$$

$$\lim_{x \rightarrow 0}{\frac{\sin(\text{sin(x)})}{\text{sin(x)}}} = 1$$

But I don't understand why.

Question: Is there an intuitive explanation for why the rule $\lim_{x \rightarrow 0}{\frac{\sin(\text{small})}{\text{same small}}} = 1$ holds?

The picture above really helped me understand the original limit, but it doesn't really help me understand why the others are true.

Answer

I'm not sure that this falls into the category of "intuitive explanation" but the general phenomenon you are considering is a consequence of the commutativity of composition of continuous functions and limits. If you let $f(x) = \frac{\sin x}x$, then for any continuous function $g$ with $g(0)=0$ we have $\lim_{x \rightarrow 0} f \circ g(x) =1$.