# Thread: Finding this limit without l'Hospital's rule

1. ## Finding this limit without l'Hospital's rule

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

We haven't covered l'Hospital's rule yet so Wolfram Alpha's not all that helpful, and even if I did know it it wouldn't have flown on the exam.

I had misread it as cosine squared x, and that is in fact how I solved it on the test which I got no credit for, for some reason.

Any hints?

2. $\lim_{x\to 0}\frac{1-cos(2x)}{x^{2}}$

$\lim_{x\to 0}\frac{1-cos(2x)}{x^{2}}\cdot\frac{1+cos(2x)}{1+cos(2x)}$

$=\lim_{x\to 0}\frac{1-cos^{2}(2x)}{x^{2}(1+cos(2x))}$

$\lim_{x\to 0}\frac{sin^{2}(2x)}{x^{2}}\cdot\lim_{x\to 0}\frac{1}{1+cos(2x)}$

$\lim_{x\to 0}\left(\frac{sin(2x)}{x}\right)^{2}\cdot \lim_{x\to 0}\frac{1}{1+cos(2x)}$

As can be seen from the left most limit, we have 4. The right one is 1/2.

And the limit is 2.

3. Hello, mbacarella!

$\lim_{x\to0} \frac{1-\cos2x}{x^2}$
Here's one way . . . probably not the shortest solution.

Multiply by $\frac{1+\cos2x}{1+\cos2x}\!:$

. . $\frac{1-\cos2x}{x^2}\cdot\frac{1+\cos2x}{1+\cos2x}$

. . $=\;\frac{1-\cos^2\!2x}{x^2(1+\cos2x)}$

. . $=\;\frac{\sin^2\!2x}{x^2(1+\cos2x)}$

. . $= \;\frac{\sin^2\!2x}{x^2}\cdot\frac{1}{1+\cos2x}$

. . $= \;4\cdot\frac{\sin^2\!2x}{4x^2}\cdot\frac{1}{1+\co s2x}$

. . $=\;4\cdot\left(\frac{\sin2x}{2x}\right)^2\cdot\fra c{1}{1+\cos2x}$

$\text{We have: }\;\lim_{x\to0}\left[4\cdot\left(\frac{\sin2x}{2x}\right)^2\cdot\frac{1 }{1+\cos2x}\right]$

. . . . . . $=\;\; 4\cdot\underbrace{\left[\lim_{x\to0}\left(\frac{\sin2x}{2x}\right)^2\right]}_{\text{This is }1^2} \cdot$ $\underbrace{\left[\lim_{x\to0}\left(\frac{1}{1+\cos2x}\right)\right]}_{\text{This is }\frac{1}{2}}$

. . . . . . $= \;\;4\cdot1\cdot\frac{1}{2}$

. . . . . . $=\;\;\boxed{2}$

Edit: too slow . . . again!
.

4. $\frac{1-\cos 2x}{x^{2}}=\frac{2\sin^{2}x}{x^{2}}.$

5. Oh! Of course!

Thank you very much.

6. If you know the series expansion it can also be done like this :

Since $\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

then

$\cos 2x = 1-\frac{4x^2}{2!}+\frac{16x^4}{4!}-\frac{64x^6}{6!}+...$

so $\frac{1-\cos 2x}{x^2} = \frac{4}{2!}-\frac{16x^2}{4!}+\frac{64x^4}{6!}+...$

taking the limit as $x \rightarrow 0$ on both sides, you get $\frac{4}{2!}=4/2=2$.