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$$\lim_{x\to0}\frac{1-\cos x}{x^2}$$

I know there are many ways to calculate this. Like L'Hopital. But for learning purposes I am not supposed to do that. Instead, I decided to do it this way:

Consider that $\cos x = 1- \sin^2 \frac{x}{2}$ (from the doulbe-angle formulas here). Therefore:

$$\frac{1-(1- \sin^2 \frac{x}{2})}{x^2} = \frac{\sin^2 \frac{x}{2}}{x^2}$$

Let us split this:

$$\frac{\sin \frac{x}{2}}{x} \cdot \frac{\sin \frac{x}{2}}{x}$$

I will just work with the left one for now. At the end, I will just multiply it by itself.

$$\frac{\sin \frac{x}{2}}{x}$$

Hm. Remember that $\frac{\sin x}{x} = 1$ when $x\to0$. If only the denominator were an $\frac{x}{2}$ instead of $x$ I could do this... Well, I can! Let's just

$$\frac{\sin \frac{x}{2}}{x}\cdot\frac{\frac{1}{2}}{\frac{1}{2}}$$

Looks a bit dumb but it kind of makes sense. This will yield

$$\frac{\color{red}{\sin \frac{x}{2}}\cdot \frac{1}{2}}{\color{red}{\frac{x}{2}}}$$

Hurray, we can now apply the formula thingy and end up with

$$\frac{\color{red}1\cdot\frac{1}{2}}{\color{red}1} = \frac{1}{2}$$

Now I need to multiply this thing by itself.

$$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

And that is the answer.

... Which is wrong. The correct answer is

$$\frac{1}{2}$$

What did I do wrong in these steps? I know many of you are able to solve this with a variety of different methods, but I'm particularly interested in this specific procedure I just tried to use. What was my mistake?

Michael Hardy
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Saturn
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3 Answers3

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You are off by a factor of two in the numerator. $\cos x = 1- \color{red}2\sin^2 \frac{x}{2}$, whereas you put $\cos x = 1- \sin^2 \frac{x}{2}$ (There might be other mistakes, I'll continue checking)

Brevan Ellefsen
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  • That's definitely it. Thank you. – Saturn Sep 27 '15 at 01:23
  • @ZolTunKul Sure thing. Quickly reading through your post, you are calculating the square root of the function and then squaring it back later, so my guess is that the function you are working with after breaking it apart will have a factor of $\sqrt{2}$ in the numerator since you will also have to take the square root of the additional factor of $2$. When you later square your answer again you will get the factor of two back in the numerator, which will yield $\frac{2}{4}$ (not $\frac{1}{4}$), which gives the result you desired – Brevan Ellefsen Sep 27 '15 at 01:25
  • That was the one. Well done. +1 – Mark Viola Sep 27 '15 at 01:37
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    To the point answer. Exactly what OP was asking for. +1 – Shailesh Sep 27 '15 at 01:53
5

$$\begin{align} \lim_{x\to 0}\frac{1-\cos x}{x^2}\frac{1+\cos x}{1+\cos x} & = \lim_{x\to 0}\frac{\sin^2x}{x^2(1+\cos x)}\\ & = \lim_{x\to 0}\frac{\sin^2x}{x^2}\lim_{x\to 0}\frac 1{1+\cos x}\\ & = 1\cdot\frac 12=\frac 12. \end{align}$$

Tim Raczkowski
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Notice, your mistake $\cos x\ne 1-\sin^2\frac{x}{2}$

Now, there are various methods to find the limit here given two methods as follows

Method-1$$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$ $$=\lim_{x\to 0}\frac{2\sin^2\left( \frac{x}{2}\right)}{x^2}=\lim_{x\to 0}\frac{1}{2}\frac{\sin^2\left( \frac{x}{2}\right)}{\left( \frac{x}{2}\right)^2}$$$$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sin\left( \frac{x}{2}\right)}{\left( \frac{x}{2}\right)}\right)^2=\frac{1}{2}(1)^2=\frac{1}{2}$$

Method-2 $$\lim_{x\to 0}\frac{1-cos x}{x^2}$$ Applying L'Hospital's rule for $\frac{0}{0}$ form $$=\lim_{x\to 0}\frac{\sin x}{2x}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}=\frac{1}{2}$$

Harish Chandra Rajpoot
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