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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?

vitamin d
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Oshawott
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    You can use L’Hôspital twice. Or you can use the small angle approximation of cos, but that’s basically the same as Taylor. – Milten Jul 11 '21 at 14:08
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    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 '21 at 14:09
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    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. – Toby Mak Jul 11 '21 at 14:32

6 Answers6

15

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

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

Toby Mak
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    (+1) This is the way I would have presented since it avoids LHR and relies on more elementary analytical tools. – Mark Viola Jul 11 '21 at 14:22
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$$\lim\limits_{x\to 0}\frac{1-\cos 7x}{3x^2}=\lim\limits_{x\to 0}\frac{1-(1-2\sin^2\frac{7x}{2})}{3x^2}=\lim\limits_{x\to 0}\frac{2\sin^2\frac{7x}{2}}{3x^2}=\frac{49}{6}.$$

Lion Heart
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5

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

José Carlos Santos
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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.

vitamin d
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3

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

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

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

$$=\frac{49}{3\cdot2}=\frac{49}{6}$$

Soumil Gupta
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2

An asymptotic approach

As $x \to 0$

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

A S D
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