Trigonometric limits without derivatives

Question

How would one derive $\lim_{x\rightarrow0}\left(\frac{\sin(x)}{x}\right)$ without using any derivatives?

Answer 1

We can proceed without using derivatives in a number of ways. We present two ways herein.

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APPROACH $1$:

First, we define the sine function as

$$\sin(x)\equiv\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!} $$

from which we see that $\frac{\sin(x)}{x}$ can be written as

$$\frac{\sin(x)}{x}=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!} \tag 1$$

The series on the right-hand side of $(1)$ converges uniformly on any closed, bounded interval. Therefore, we have

$$\begin{align} \lim_{x\to 0}\frac{\sin(x)}{x}&=\lim_{x\to 0}\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}\\\\ &=\sum_{n=0}^\infty \,\lim_{x\to 0}\left(\frac{(-1)^nx^{2n}}{(2n+1)!}\right)\\\\ &=1 \end{align}$$

APPROACH $2$:

Second, we define the sine function as the inverse function defined by the integral

$$\arcsin(x)\equiv \int_0^x \frac{1}{\sqrt{1-t^2}}\,dt \tag 2$$

for $|x|<1$. From $(2)$, we see that both the arcsine function and its inverse are continuous. Moreover, it is easy to see that $\arcsin(0)=0$, thereby implying that $\sin(0)=0$.

Next, enforcing the substitution $x\to \sin(x)$ in $(2)$ reveals

$$x=\int_0^{\sin(x)}\frac{1}{\sqrt{1-t^2}}\,dt \tag 3$$

Dividing $(3)$ by $\sin(x)$ reveals

$$\frac{x}{\sin(x)}=\frac{1}{\sin(x)}\int_0^{\sin(x)}\frac{1}{\sqrt{1-t^2}}\,dt \tag 4$$

Appealing to the mean value theorem for integrals, we see that there exists a number $\xi \in (0,\sin(x))$ such that

$$\frac{x}{\sin(x)}=\frac{1}{\sqrt{1-\xi^2}}$$

Then, since $\sin(x)$ is continuous with $\sin(0)=0$, we find using the squeeze theorem

$$\lim_{x\to 0}\frac{x}{\sin(x)}=\lim_{x\to 0}\frac{1}{\sqrt{1-\xi^2}}=1$$

which implies

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

as expected.