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?

• You can use L’Hôspital twice. Or you can use the small angle approximation of cos, but that’s basically the same as Taylor. Jul 11 at 14:08
• You can write $$\frac{{1 - \cos (7x)}}{{3x^2 }} = \frac{{49}}{6}\left( {\frac{{\sin \left( {\frac{7}{2}x} \right)}}{{\frac{7}{2}x}}} \right)^2$$ using the known identity $\frac{{1 - \cos w}}{2} = \sin ^2 \left( {\frac{w}{2}} \right)$.
– Gary
Jul 11 at 14:09
• Alternative, this is the real part of $\lim_{z \to 0} \frac{1 - e^{7iz}}{3z^2} = \lim_{z \to 0} \frac{1 - (1 + 7iz - 49z^2/2 + \cdots)}{3z^2} = \frac{49}{6}$. The series expansion works when $z$ is complex because $e^z$ is an analytic function. Jul 11 at 14:32

Multiplying top and bottom by $$1 + \cos 7x$$:

$$\lim_{x\to 0}\frac{1-\cos 7x}{3x^2} = \lim_{x\to 0}\frac{(\sin 7x)^2}{3x^2} \frac{1}{1 + \cos 7x} = \lim_{x\to 0}\frac{(\sin 7x)^2}{(7x)^2} \frac{49/3}{1 + \cos 7x} = 1 \cdot \frac{49/3}{2} = \boxed{\frac{49}{6}}.$$

• (+1) This is the way I would have presented since it avoids LHR and relies on more elementary analytical tools. Jul 11 at 14:22
• This way seems the most natural to me. I don't think we need anything more than this. Jul 11 at 14:23

$$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}=\lim\limits_{x\to 0}\frac{1-(1-2\sin^2\frac{7x}{2})}{3x^2}=\lim\limits_{x\to 0}\frac{2\sin^2\frac{7x}{2}}{3x^2}=\frac{49}{6}.$$

• (+1) for applying elementary analytical tools only. Jul 11 at 14:24

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 $$\lim_{x\to0}\frac{\sin(x)}x=1$$.

• Why the down vote? Jul 11 at 14:16
• Exactly, I don't see why L'Hopital's can't be used. Jul 11 at 14:16
• This looks like the simplest solution to me. Jul 11 at 14:17
• @Unknown: L'Hôpital's rule states that if $\lim_{x \to 0}f(x)=\lim_{x \to 0}g(x)=0$, then $$\lim_{x \to 0}\frac{f(x)}{g(x)}=\lim_{x \to 0}\frac{f'(x)}{g'(x)} \, ,$$ provided that the limit on the RHS exists.
– Joe
Jul 11 at 14:29
• For the final limit, we can make the substitution $u=7x$:$$\lim_{x \to 0}\frac{7\sin(7x)}{6x}=\lim_{u \to 0}\frac{49\sin(u)}{6u}=\frac{49}{6}\cdot\lim_{u\to0}\frac{\sin u}{u}=\frac{49}{6} \cdot 1=\frac{49}{6} \, .$$
– Joe
Jul 11 at 14:33

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.

• $$(1-\cos 7x)' = 7\sin 7x$$ not $7\sin x$ Jul 11 at 14:19
• Sorry for the typo, now corrected. (At least in the limit it was correct.) Jul 11 at 14:19
• LHR is a bit overkill ;-) Jul 11 at 14:23

it is already known that: $$\lim\limits_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$ Now Manipulating,

$$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}.\frac{49}{49}$$

=$$\lim\limits_{x\to 0}\frac{1-\cos 7x}{(7x)^2}.\frac{49}{3}.$$

= $$\frac{49}{3 X 2}$$ =49/6

An asymptotic approach

As $$x \to 0$$

$$\frac{1-\cos 7x}{3x^2} \sim \frac{49x^2/2}{3x^2}=\frac{49}{6}$$