2
$\begingroup$

I've been told to solve the following limit using only $\lim_{x \to 0}\frac{\sin x}{x}=1$:

$$\lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4}$$

I don't know how to do it except using L'Hospital's Rule but it's just insane amount of math and not the writer's intention(Symbolab do it but it's a computer so ;)

The answer is: $\frac18$

I'm looking for a way to solve this using $\frac{\sin x}x$.

$\endgroup$
7
$\begingroup$

Appealing to the half-angle identity, $1-\cos(x)=2\sin^2(x/2)$, reveals

$$\frac{1-\cos(1-\cos(x))}{x^4}=\frac{2\sin^2(\sin^2(x/2))}{x^4}$$

Next, we write

$$\begin{align} \frac{2\sin^2(\sin^2(x/2))}{x^4}&=2\left(\frac{\sin(\sin^2(x/2))}{x^2}\right)^2\\\\ &=2\left(\frac{\sin(\sin^2(x/2))}{\sin^2(x/2)}\,\frac{\sin^2(x/2)}{x^2}\right)^2\\\\ &=2\,\color{blue}{\underbrace{\left(\frac{\sin^2(\sin^2(x/2))}{\sin^2(x/2)}\right)^2}_{\to 1}}\,\color{red}{\underbrace{\frac1{16}\left(\frac{\sin^2(x/2)}{(x/2)^2}\right)^2}_{\to 1/16}}\\\\ &\to \frac18 \end{align}$$

$\endgroup$
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Aug 10 '17 at 18:56
4
$\begingroup$

$$\begin{array}{rcl} \displaystyle \lim_{x \to 0} \frac{1-\cos(1-\cos x)}{x^4} &=& \displaystyle \lim_{x \to 0} \frac{1-\cos^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\ &=& \displaystyle \lim_{x \to 0} \frac{\sin^2(1-\cos x)}{x^4[1+\cos(1-\cos x)]} \\ &=& \displaystyle \lim_{x \to 0} \frac{(1-\cos x)^2\sin^2(1-\cos x)}{x^4(1-\cos x)^2[1+\cos(1-\cos x)]} \\ &=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos x)^2}{x^4[1+\cos(1-\cos x)]} \\ &=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(1-\cos^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\ &=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2\frac{(\sin^2 x)^2}{x^4(1+\cos x)^2[1+\cos(1-\cos x)]} \\ &=& \displaystyle \lim_{x \to 0} \left(\frac{\sin(1-\cos x)}{1-\cos x}\right)^2 \left(\frac{\sin x}{x}\right)^4 \frac{1}{(1+\cos x)^2[1+\cos(1-\cos x)]} \\ &=& 1^2 \cdot 1^4 \cdot \dfrac{1}{2^2 \cdot 2} \\ &=& \dfrac18 \end{array}$$

$\endgroup$
2
$\begingroup$

Don't forget $1-\cos x=2\sin^2(x/2)$. Therefore $$1-\cos(1-\cos x)=2\sin^2\left(\sin^2\frac x2\right) $$ and note $$\frac{\sin(\sin^2 x/2)}{\sin^2 x/2}\to1$$

$\endgroup$
  • $\begingroup$ You have a factor of $2$ that doesn't belong. Note $2\sin^2\left(\frac{1-\cos(x)}{2}\right)=2\sin^2(\sin^2(x/2))$ $\endgroup$ – Mark Viola Aug 10 '17 at 17:34
  • $\begingroup$ There you go ... just like mine ... ;-)) $\endgroup$ – Mark Viola Aug 10 '17 at 17:40
1
$\begingroup$

As $u\to 0,$

$$\frac{1-\cos u}{u^2} = \frac{1}{1+\cos u}\frac{1-\cos^2 u}{u^2} = \frac{1}{1+\cos u}\frac{\sin^2 u }{u^2}\to \frac{1}{2}\cdot 1^2 = \frac{1}{2}.$$

Therefore as $x\to 0,$

$$\frac{1-\cos (1-\cos x)}{x^4}= \frac{1-\cos (1-\cos x)}{(1-\cos x)^2}\frac{(1-\cos x)^2}{x^4}$$ $$ = \frac{1-\cos (1-\cos x)}{(1-\cos x)^2}\left (\frac{1-\cos x}{x^2}\right)^2 \to \frac{1}{2}\cdot \left ( \frac{1}{2}\right) ^2 = \frac{1}{8}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.