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?
 A: $$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}=\lim\limits_{x\to 0}\frac{1-(1-2\sin^2\frac{7x}{2})}{3x^2}=\lim\limits_{x\to 0}\frac{2\sin^2\frac{7x}{2}}{3x^2}=\frac{49}{6}.$$
A: By L'Hopital's Rule,$$\lim_{x\to0}\frac{1-\cos(7x)}{3x^2}=\lim_{x\to0}\frac{7\sin(7x)}{6x},$$if this last limit exists. Which it does. You can apply L'Hopital's Rule a second time, or use the fact that $\lim_{x\to0}\frac{\sin(x)}x=1$.
A: Use l'hopital. The derivative of the numerator is $(1-\cos 7x)'=7\sin 7x$ and of the denominator $(3x^2)=6x$. We get
$$\lim_{x\to0}\frac{7\sin(7x)}{6x}=\lim_{x/7\to0}\frac{7\sin(x)}{6(x/7)}=\lim_{x\to0}\frac{49\sin(x)}{6x}=\frac{49}{6}.$$
In the second step we used the substitution $x\mapsto x/7$, see here why it's justified.
A: It is already known that:
$$\lim\limits_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$
Now Manipulating,
$$=\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.\frac{49}{49}$$
$$=\lim\limits_{x\to 0}\frac{1-\cos 7x}{(7x)^2}.\frac{49}{3}$$
$$=\frac{49}{3\cdot2}=\frac{49}{6}$$
A: An asymptotic approach
As $x \to 0$
$$\frac{1-\cos 7x}{3x^2} \sim \frac{49x^2/2}{3x^2}=\frac{49}{6}$$
A: Multiplying top and bottom by $1 + \cos 7x$:
$$ \lim_{x\to 0}\frac{1-\cos 7x}{3x^2} = \lim_{x\to 0}\frac{(\sin 7x)^2}{3x^2} \frac{1}{1 + \cos 7x} = \lim_{x\to 0}\frac{(\sin 7x)^2}{(7x)^2} \frac{49/3}{1 + \cos 7x} = 1 \cdot \frac{49/3}{2}  = \boxed{\frac{49}{6}}.$$
