Another proof of $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$

Question

I am sorry if this is a duplicate question, but as far as I searched I have not come across this question.

$\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$. This formula is famously proven by geometrical means using area of a circle and so on..

I want to know if this method of proof over l-hopital's Rule is also acceptable. $$\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta} = \lim_{\theta \to 0} \frac{\cos \theta}{1} = 1$$ If there is any other method to prove apart from the two methods above. Please Share. Thank You!

Answer 1

Hint $:$ $\sin x < x < \tan x$ for $x \in \left (0, \frac {\pi} {2} \right )$ and $\tan x < x < \sin x$ for $x \in \left (-\frac {\pi} {2} , 0 \right ).$ Now use Squeeze theorem to conclude.

Answer 2

$$sin(\theta)=\sum_{n=0}^\infty \frac{(-1)^n{\theta}^{2n+1}}{(2n+1)!}=\theta-\frac{(\theta)^3}{3!}+\frac{(\theta)^5}{5!}+....$$then$$\frac{sin(\theta)}{\theta}=1-\frac{(\theta)^2}{3!}+\frac{(\theta)^4}{5!}+....$$ and the limit as $\theta\rightarrow 0$ is 1

Answer 3

An other proof using the definition of the derivative, actually it is equivalent to the Hospital Rule but you explain why this result works. We have on one hand $\sin'(0)=\cos(0)=1$. On the other hand $\sin'(0) =\lim\limits_{\theta \to 0} \frac{\sin\theta - \sin 0}{\theta - 0} =\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta}$. So $\lim\limits_{\theta \to 0} \frac{\sin\theta}{\theta}=1$.