Evaluate following limit: $\lim_{x \to 0}\frac{\sin x}{x}$

Question

I am asked to solve following limit problem:

$\lim_{x \to 0}\frac{\sin x}{x}$

It is pretty straightforward using L'Hôpital's rule:

Since $$ \lim_{x \to 0}\frac{\sin x}{x} = \frac{0}{0}$$ Then we take derivative of the numerator and denominator, getting: $$ \lim_{x \to 0}\frac{\cos x}{1} =1$$

However, I would like to know whether there is a way to solve the limit problem above without L'Hôpital's rule. Is there?

Do you know that $\sin , x$ has a Taylor series expansion around $0$? — Kavi Rama Murthy, Aug 12 '19 at 09:56
Yes, that can be proved without L'Hopital. In fact you'd better, since that limit is usually what you need to find the derivative of $\sin$ in the first place. So using L'Hopital is circular resoning. — Ethan Bolker, Aug 12 '19 at 09:57
@Nelver: A Google search on "limit sinx/x" yields a number of links to proofs. — quasi, Aug 12 '19 at 10:00
@Nelver: A search of Math.SE for "proof limit sin x/x" yields almost a thousand results. (Not all are specifically about this limit, of course.) I suspect you can find something useful among them. — Blue, Aug 12 '19 at 10:03

score 1 · Answer 1 · answered Aug 12 '19 at 09:55

1

You could find the proof in any calculus textbook. Or, here, for instance.

https://en.wikipedia.org/wiki/Squeeze_theorem#Second_example

answered Aug 12 '19 at 09:55

+1, although usually link only answers are discouraged on MSE. – Ethan Bolker Aug 12 '19 at 09:59
Thank you for letting me know! – Aug 12 '19 at 10:02

Evaluate following limit: $\lim_{x \to 0}\frac{\sin x}{x}$

1 Answers1