Limit of $\frac{1-\cos(x)(\cos(2x))^2}{x^2}$ when $x\to0$ without L'Hôpital's rule?

Question

I need help to compute this limit without L'Hôpital's rule:

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

Answer 1

Hint:

$$\lim_{x\to 0}\frac{1-\cos(x)(\cos(2x))^2}{x^2}=$$ $$ =\lim_{x\to 0}\dfrac{\cos^2(2x)+\sin^2(2x)-\cos(x)\cos^2(2x)}{x^2}= $$ $$ =\lim_{x\to 0}\dfrac{\cos^2(2x)(1-\cos(x))+\sin^2(2x)}{x^2}= $$ $$ \lim_{x\to 0}\left[\cos^2(2x)\dfrac{1-\cos(x)}{x^2}+4\dfrac{\sin(2x)}{2x}\dfrac{\sin(2x)}{2x} \right] $$

Now use the notable limits (that can be proved without L'Hpital) :

$$ \lim_{x\to 0}\dfrac{1-\cos(x)}{x^2}=\dfrac{1}{2} \qquad \lim_{x\to 0}\dfrac{\sin(x)}{x}=1 $$ and find $\lim_{x \to 0} f(x)=\frac{9}{2}$

Answer 2

I assume you know at least that $\lim_{x\to0}\frac{\sin x}{x}=1$.

You have the identity $\cos (2x)=1-2\sin^2x$.

$$1-\cos(x)(\cos(2x))^2=1-(1-2\sin^2 (x/2))(1-2\sin^2x)^2$$ $$=1-\left[1-2\sin^2(x/2)-2\sin^2(x)+4\sin^2(x/2)\sin^2(x)\right](1-2\sin^2 x)$$ $$=\left[2\sin^2(x/2)+2\sin^2(x)-4\sin^2(x/2)\sin^2(x)\right](1-2\sin^2(x))+2\sin^2x$$

Thus

$$\frac{1-\cos(x)(\cos(2x))^2}{x^2}=\left[2\frac{\sin^2(x/2)}{x^2}+2\frac{\sin^2(x)}{x^2}-4\sin^2(x/2)\frac{\sin^2(x)}{x^2}\right](1-2\sin^2(x))+2\frac{\sin^2x}{x^2}$$

Now, for the terms inside the brackets,

$$2\frac{\sin^2(x/2)}{x^2}=\frac12\left(\frac{\sin(x/2)}{(x/2)}\right)^2\underset{x\to0}{\longrightarrow}\frac12$$

$$2\frac{\sin^2(x)}{x^2}=2\left(\frac{\sin(x)}{x}\right)^2\underset{x\to0}{\longrightarrow}2$$

And the third term has limit $0$, since you have one factor that tends to $1$, and the other to $0$: $$\frac{\sin^2(x/2)\sin^2(x)}{x^2}=\sin^2(x/2)\left(\frac{\sin(x)}{x}\right)^2\underset{x\to0}{\longrightarrow}0$$

The $(1-2\sin^2x)$ factor tends also to $1$, so the limit of the first term (the big thing with brackets $\times(1-2\sin^2x)$) has limit $5/2$.

The last term has limit:

$$2\frac{\sin^2x}{x^2}=2\left(\frac{\sin x}{x}\right)^2\underset{x\to0}{\longrightarrow}2$$

All in all, your limit is $5/2+2=9/2$.

Answer 3

My first impulse for problems like this is to get everything into one "level" of trig function, i.e., replace $\cos2x$ with $2\cos^2x-1$. Doing a bit of algebra, one obtains

$$1-\cos x\cos^22x=1-\cos x(2\cos^2x-1)^2=(1-\cos x)(1+4\cos^3x+4\cos^4x)$$

This gives

$$\lim_{x\to0}{1-\cos x\cos^22x\over x^2}=\lim_{x\to0}{1-\cos x\over x^2}\cdot\lim_{x\to0}(1+4\cos^3x+4\cos^4x)=\lim_{x\to0}{1-\cos x\over x^2}\cdot9$$

Finally,

$$1-\cos x={(1-\cos x)(1+\cos x)\over1+\cos x}={1-\cos^2x\over1+\cos x}={\sin^2x\over1+\cos x}$$

so

$$\lim_{x\to0}{1-\cos x\over x^2}=\left(\lim_{x\to0}{\sin x\over x} \right)^2\lim_{x\to0}{1\over1+\cos x}={1\over2}$$

Putting everything together gives the overall limit $9/2$.

An important point here is that it's often possible to use the multiplicative property $\lim(fg)=\lim(f)\lim(g)$ (when all three limits exist) to peel off pieces of a limit that simply amount to a straightforward function evaluation.

Remark: I should note that my "first impulse" doesn't always work; sometimes it just leads to a big mess, and I have to try something else. Nor does it produce the nicest solution. There's a lot to like here in Emilio Novati's answer.

Answer 4

We can proceed as follows

$$\begin{align} \frac{1-\cos^2(2x)\cos x}{x^2}&=\frac{1-\cos^2(2x)}{x^2}+\cos^2(2x)\frac{1-\cos x}{x^2}\\\\ &=4\left(\frac{\sin(2x)}{2x}\right)^2+\frac12\cos^2(2x)\left(\frac{\sin(x/2)}{x/2}\right)^2 \end{align}$$

where we used the identity $1-\cos z=2\sin^2(z/2)$.

Now use $\lim_{z\to 0}\frac{\sin z}{z}=1$.