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

Question

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 1

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 $\lim_{x \rightarrow 0} f \circ g(x) =1$.

Answer 2

This is because

• $\lim_{x\to a}f(x)=b$

is equivalent to

• $\lim_{n\to\infty}f(x_n)=b$ for every sequence $(x_n)_{n\in\Bbb N}$ with $x_n\to a$.

Answer 3

So we know (by know here I mean we know the proof) that the following is true $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ Now the symbol $x$ used here is a variable and it does not matter what value or quantity it represents as long as it tends to $0$. We can by a certain abuse of notation write the same equation in a crude manner as $$\lim_{\text{"small thing" tends to zero}}\frac{\sin(\text{"small thing"})}{\text{"small thing"}} = 1$$ In more formal terms let $f$ be a function of $x$ such that $$\lim_{x \to a}f(x) = 0\tag{2}$$ and $$f(x) \neq 0\tag{3}$$ as $x \to 0$. Then we can write the equation $(1)$ as $$\lim_{x \to a}\frac{\sin f(x)}{f(x)} = 1\tag{4}$$ whenever equations $(2), (3)$ hold.

That equations $(1), (2), (3)$ lead to equation $(4)$ is proved via the use of rule of substitution. The important idea to note is that $(\sin x)/x$ tends to $1$ as $x \to 0$ and it does not depend on the manner in which $x \to 0$. Thus for example $x^{2}$ also tends to $0$ as $x \to 0$ but it tends much faster and hence $(\sin (x^{2}))/x^{2}$ tends to $1$. Equation $(4)$ says the same thing that it does not matter how slow / fast / or in whatever manner the argument $f(x)$ of $\sin $ tends to $0$ the expression $(\sin f(x))/f(x)$ tends to $1$ as long as we ensure that $f(x)$ never actually becomes $0$ (because then the expression would be meaningless due to division by $0$). So the intuitive explanation of $(4)$ is ultimately dependent on intuitive explanation of $(1)$ (which you already mention in your question).

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I am not sure why there is a prevalent idea that for equation $(4)$ to hold we must have $f$ continuous at $x = a$ (this is also mentioned in the accepted answer). This is not necessary at all. The "small thing" mentioned in the beginning is only supposed to tend to $0$ and not that it should tend to $0$ in a continuous manner.