# Find the limit $\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$

Find the limit $\lim_{x\to 0}\frac{1-\cos 2x}{x^2}$

This is what I did: $\lim_{x\to 0}\frac{1-\cos 2x}{x^2} = \frac{0}{0}$

Then, if we apply L'hopital's, we get: $\lim_{x\to 0}\frac{2\sin 2x}{2x} =\frac{0}{0}$.

Once again, using L'hopital's rule, we get:$\lim_{x\to 0}\frac{4\cos 2x}{2} = 2\cos 2x = 2$.

Can someone please tell me what I did wrong here? Thanks.

Update: Thanks everyone for your wonderful answers. I have found out the reason for taking a point off of my work. It is because I didn't use the correct expression. For example, since lim$_{x\to 0} \frac{f(x)}{g(x)} = \frac{0}{0},$ I didn't write the correct term "lim$_{x \to 0} \frac{f'(x)}{g'(x)}" = ...$ and instead I equated everything when I was applying L'Hapital's rule. So, I thought I should've mention it here. Thanks again.

• As far as I can tell, you're entirely correct. – JacksonFitzsimmons Jul 12 '15 at 19:29
• @JacksonFitzsimmons could you please look even more closely because this is the exact procedure I did for my hw but still a point taken off. So, maybe I did something wrong here. – Jellyfish Jul 12 '15 at 19:32
• Look at Ivo's answer, you probably lost a point for applying L'hopital's rule when you didn't have to. – JacksonFitzsimmons Jul 12 '15 at 19:34
• Strictly speaking, it is not correct to say $\lim\limits_{x\to0}\dfrac{1-\cos(2x)}{x^2} = \dfrac 0 0$. In this case the limit is an actual number, not $\dfrac 0 0$. But one can say the indeterminate form with which one is dealing is "$\dfrac 0 0$". ${}\qquad{}$ – Michael Hardy Jul 12 '15 at 19:50
• Various things are not well expressed. For example you write $\lim_{x\to 0}\frac{4\cos 2x}{2}=2\cos 2x=2$. You should have written $\lim_{x\to 0}\frac{4\cos 2x}{2}=\lim_{x\to 0} 2\cos 2x=2$. – André Nicolas Jul 12 '15 at 20:17

You are correct, but there is no need for a second L'Hospital, because: $$\lim_{x \to 0}2\frac{\sin 2x}{2x} = 2 \lim_{x \to 0}\frac{\sin 2x}{2x} = 2\lim_{u \to 0}\frac{\sin u}{u} = 2 \cdot 1 =2.$$

• Sorry, if it's a stupid question. But, isn't as lim$_{u\to 0} \frac{sin u}{u} = 0$? – Jellyfish Jul 12 '15 at 19:45
• No, it is $1$. This is one of the fundamental limits. This is not a proof, but you can found out the result using L'Hospital again (which is circular!): $$\lim_{u \to 0}\frac{\sin u}{u} = \lim_{u \to 0}\cos u = \cos 0 = 1.$$ – Ivo Terek Jul 12 '15 at 19:49
• Thanks, I don't know how I missed that. But, we still have to use second L'Hopital in that case. – Jellyfish Jul 12 '15 at 20:00
• This limit is provable without L'Hospital. In fact, you use that limit to prove that $\sin'x = \cos x$, which in turn allows you to use L'Hospital. That why I said it was circular. – Ivo Terek Jul 12 '15 at 20:02
• You can use the fact that for $x$ close to, but not equal to $0$, $\cos(x) < \frac{\sin(x)}{x} < 1$, and apply squeeze theorem at $0$. – Eemil Wallin Jul 13 '15 at 0:56

Without L'Hopital: Multiply top and bottom by $1+\cos 2x$ to get

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

It's easily seen that $(\sin^2 2x)/x^2 \to 4.$ Thus the limit is $4\cdot [1/(1+1)] = 2.$

Alternatively, your limit can be written as $\lim_{x \to 0} \frac{1- (\cos^{2}x-\sin^{2}x)}{x^{2}} = \lim_{x \to 0} \frac{2\sin^{2}x}{x^{2}}$. Since $\lim_{x \to 0} \frac{ \sin x}{x} = 1$, ( a standard limit which might have been given in your course already- you could use L'Hopital, but you really need to know this limit to prove that $\sin x$ is differentiable at $0$ anyway), your limit is $2$.

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

$$\lim_{x\to 0} \left(\frac{\frac{d}{dx}\sin(2x)}{\frac{d}{dx}x}\right)=$$ $$\lim_{x\to 0} \left(\frac{2\cos(2x)}{1}\right)=$$ $$\lim_{x\to 0} \left(2\cos(2x)\right)=$$

(Since $2\cos(2x)$ is a continuous function of $x$):

$$2\cos(2\times0)=2\cos(0)=2\times1=2$$

$$\lim_{x\to 0} \left(\frac{1-\cos(2x)}{x^2}\right) = \lim_{x\to 0} \left(\frac{1-\cos(2x)}{(2x)^2}\frac{4x^2}{x²}\right) =\frac{1}{2}\times 4 = 2$$

• I didn't see that it was shared 3 years ago ,my bad . But i just gave my answer , it could help as long as it's a different answer from those previously given . – LeFo Oct 3 '18 at 15:01