TITLE: How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$?
QUESTION [6 upvotes]: Evaluate $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.$$

I solved the problem with the Taylor series expansion of $\cos x$. Here is my solution:
$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1-\{1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}-\frac{(7x)^6}{6!}+\dots\}}{3x^2}\\ =\lim\limits_{x\to 0}\frac{x^2(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)}{3x^2}\\ =\lim\limits_{x\to 0}\frac{1}{3}(\frac{7^2}{2!}-\frac{7^4x^2}{4!}+\frac{7^6x^4}{6!}-\dots)\\ =\frac {49}{6}$
Can this be solved without using the Taylor series?

REPLY [2 votes]: An asymptotic approach
As $x \to 0$
$$\frac{1-\cos 7x}{3x^2} \sim \frac{49x^2/2}{3x^2}=\frac{49}{6}$$