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