How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$

Question

How can I show that

$$ \lim_{x\to 0} \frac{1-\cos^n x}{x^2} = \frac{n}{2} $$

without using Taylor series $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?

This is one of the cases where l'Hospital's rule also works well. Is that allowed? Also, $1 - \cos^n x = (1-\cos x)(1 + \cos x + \cos^2 x + \dotsc + \cos^{n-1} x)$. — Daniel Fischer, Mar 08 '16 at 18:38
@Arthur Hence the question whether it's allowed. Some people are of the opinion that l'Hospital's rule is "less advanced" than Taylor expansions, and if the author of the exercise belongs to that group, l'Hospital might be the intended way. — Daniel Fischer, Mar 08 '16 at 18:42
$1-cos^n x= (1-cos x)(...) $ and $\frac{1-cosx}{x^2} \to \frac{1}{2}$ is good for me. Thanks a lot! — Asuaya Iora, Mar 08 '16 at 18:53

score 4 · Accepted Answer · answered Mar 08 '16 at 18:43

4

From l'Hopital's rule ... $$L = \lim_{x\to 0}\frac{1-\cos^n x}{x^2} = \lim_{x\to 0}\frac{n\sin x\cos^{n-1}x}{2x} = \lim_{x\to 0}\frac n2\left(\frac{\sin x}x\right)\left(\cos^{n-1}x\right) = \cdots$$

answered Mar 08 '16 at 18:43

jameselmore

5,207

score 3 · Answer 2 · answered Mar 08 '16 at 19:19

Here, we use an approach that is more efficient and more elementary than use of L'Hospital's Rule.

Simply factor the term $1-\cos^n(x)$ as

$$1-\cos^n(x)=(1-\cos(x))\sum_{m=0}^{n-1}\cos^m(x)$$

Then, we have

$$\begin{align} \lim_{x\to 0}\frac{1-\cos^n(x)}{x^2}&=\lim_{x\to 0}\frac{1-\cos(x)}{x^2}\sum_{m=0}^{n-1}\cos^m(x)\\\\ &\lim_{x\to 0}\frac{1-\cos(x)}{x^2} \lim_{x\to 0} \sum_{m=0}^{n-1}\cos^m(x)\\\\ &=\left(\frac12\right)(n)\\\\ &=\frac n2 \end{align}$$

score 2 · Answer 3 · answered Mar 09 '16 at 04:45

I have seen several limit problems on MSE where people don't use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ whereas frequent use is made of other standard limits like $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{2}$$ and this question is also an instance where the limit $(1)$ should be used.

We have \begin{align} L &= \lim_{x \to 0}\frac{1 - \cos^{n}x}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{1 - \cos^{n}x}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{1 - \cos^{n}x}{1 - \cos x}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\notag\\ &= \lim_{t \to 1}\frac{1 - t^{n}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos^{2}x}{x^{2}(1 + \cos x)}\text{ (putting }t = \cos x)\notag\\ &= \lim_{t \to 1}\frac{t^{n} - 1}{t - 1}\cdot\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \cos x}\notag\\ &= n\cdot 1\cdot\frac{1}{1 + 1}\notag\\ &= \frac{n}{2}\notag \end{align}

score 1 · Answer 4 · answered Mar 08 '16 at 18:44

$$\lim_{x\to0}\frac{1-\cos^n(x)}{x^2}=$$

Applying l'Hôpital's rule, we get that:

$$\lim_{x\to0}\frac{\frac{\partial}{\partial x}\left(1-\cos^n(x)\right)}{\frac{\partial}{\partial x}\left(x^2\right)}=\lim_{x\to0}\frac{n\sin(x)\cos^{n-1}(x)}{2x}=$$

By the product rule:

$$\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]\left[\lim_{x\to0}\cos^{n-1}(x)\right]=\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]\left[\cos^{n-1}(0)\right]=$$ $$\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]\left[1\right]=\frac{1}{2}\left[\lim_{x\to0}\frac{n\sin(x)}{x}\right]=\frac{1}{2}\left[n\lim_{x\to0}\frac{\sin(x)}{x}\right]=$$

Applying l'Hôpital's rule, we get that:

$$\frac{1}{2}\left[n\lim_{x\to0}\frac{\frac{\partial}{\partial x}\left(\sin(x)\right)}{\frac{\partial}{\partial x}\left(x\right)}\right]=\frac{1}{2}\left[n\lim_{x\to0}\frac{\cos(x)}{1}\right]=\frac{1}{2}\left[n\lim_{x\to0}\cos(x)\right]=$$ $$\frac{1}{2}\left[n\cos(0)\right]=\frac{1}{2}\left[n\cdot1\right]=\frac{1}{2}\left[n\right]=\frac{n}{2}$$

To apply L'Hopital to compute $\lim\limits_{x\to0}\frac{\sin x}x$ when $x\to0$ is a circular reasoning, well documented on the site. The reason is that $\lim\limits_{x\to0}\frac{\sin x}x$ is, by definition, the derivative of the sine at zero. So, if one knows that this derivative is the cosine (and if one knows the value of $\cos 0$), one already knows the limit. (The first instance of L'Hopital in this answer is probably the most complicated way to proceed, but at least it is not logically flawed like the second instance is.) — Did, Mar 10 '16 at 06:35

score 0 · Answer 5 · answered Mar 08 '16 at 20:43

0

You just need to know that $\lim_{x\to 0}\frac{\sin x}{x}=1$, since:

$$\begin{eqnarray*} \frac{1-\cos^n x}{x^2} &=& \frac{1-\cos x}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x) = \frac{2\sin^2\frac{x}{2}}{x^2}\cdot \sum_{k=0}^{n-1}\cos^k(x)\\ &=& \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \cdot \sum_{k=0}^{n-1}\cos^k(x)\stackrel{x\to 0}{\longrightarrow}\color{red}{\frac{n}{2}}.\end{eqnarray*}$$

answered Mar 08 '16 at 20:43

Jack D'Aurizio

353,855

This looks a bit familiar ... ;-)) ... nice to see you back Jack! - Mark – Mark Viola Mar 09 '16 at 04:52

score 0 · Answer 6 · 2016-03-08T21:10:09.657

0

Without L'Hospital:

Thanks to $\dfrac{\sin(x)}x\to1$, one can interchange $x$ and $\sin(x)$ and rewrite the limits as

$$\lim_{x\to0}\frac{1-(1-x^2)^{n/2}}{x^2}=\lim_{x\to0}\frac{1-(1-x^2)^n}{x^2(1+(1-x^2)^{n/2})}.$$

And by the binomial theorem,

$$ \lim_{x\to0}\frac{1-1+nx^2-\binom n2x^4+\binom n3x^6\cdots}{2x^2}=\frac n2.$$

edited Mar 08 '16 at 21:10

answered Mar 08 '16 at 20:53

How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$

6 Answers6