How to evaluate $\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}$? 
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?

It is already known that:
$$\lim\limits_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$
Now Manipulating,
$$=\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.\frac{49}{49}$$
$$=\lim\limits_{x\to 0}\frac{1-\cos 7x}{(7x)^2}.\frac{49}{3}$$
$$=\frac{49}{3\cdot2}=\frac{49}{6}$$
