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

Question

Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$

The answer stated is $\displaystyle {1 \over 6}$.

What I've tried: $$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$

$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x) -1+1- \cos x}{x^4}$$

$$=\displaystyle \lim_{x\to0} \frac{1- \cos x}{x^4} - \frac {1-\cos (\sin x)}{x^4}$$

$$=\displaystyle \lim_{x\to0} \frac{2 \sin^2(\frac {x}{2})}{x^4} - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$

$$=\displaystyle \lim_{x\to0} \left(\frac{\sin(\frac {x}{2})}{x} \right)^2. \left( \dfrac{1}{2x^2} \right) - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$

I'm not sure how I can evaluate the limit by proceeding this way. All help will be appreciated.

P.S. I'd prefer not using L'Hôpital's rule, it can get really messy.

EDIT: I should have mentioned that I would prefer if the solution does not use taylor series approximations (or any approximations) for that matter.

L'hopital may be messy but it should get the work done, since you know that after 4 derivatives, x^4 will become a constant. — Julian Mejia, May 13 '19 at 16:18
What's wrong with using L'Hôpital's rule? applying it twice may be enough to simplify this limit — Vasili, May 13 '19 at 16:19
Without using l'hopital i would try expand it using a taylor series, if you do this correctly youll get 1/6 — ricky, May 13 '19 at 16:23
Possible duplicate of Calculating limit of function. Or $\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of n such that l is non zero finite real number. Find l (especially siméon's answer for a proof without l'hopital's rule) — YuiTo Cheng, May 13 '19 at 16:33
I should mention that I'd prefer an answer without using taylor series expansions; but all 3 answers below use taylor series. I'm okay with it, but I was looking for a way that does not involve approximations/differenciation. I found a satisfactory solution here. — QuasarChaser, May 13 '19 at 17:08
https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion — lab bhattacharjee, May 13 '19 at 18:41

score 5 · Answer 1 · answered May 13 '19 at 16:57

By using trigonometry identity and Taylor series,\begin{align} &\lim_{x\to0} \frac{\cos (\sin x)-\cos(x)}{x^4}\\ &=-2\lim_{x\to 0}\frac{\sin \left( \frac{\sin x - x}2\right) \sin\left(\frac{\sin x + x}2 \right)}{x^4}\\ &= -2\lim_{x \to 0 }\frac{\sin \left( \frac{-x^3}{2(6)}\right)\sin \left( \frac{x+x}{2}\right)}{x^4}\\ &= -2 \lim_{x \to 0}\frac{-\frac{x^3}{6(2)}\cdot x}{x^4}\\ &= \frac16\end{align}

score 3 · Answer 2 · answered May 13 '19 at 16:27

Note that from the expansions $\sin(x)=x-\frac{x^3}{6}+O(x^5)$ and $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have

$$\begin{align} \cos(\sin(x))&=\cos\left(x-\frac{x^3}{6}+O(x^5)\right)\\\\ &=1-\frac12\left(x-\frac{x^3}{6}+O(x^5)\right)^2+\frac{1}{24}\left(x-\frac{x^3}{6}+O(x^5)\right)^4+O(x^6)\\\\ &=1-\frac12x^2+\frac{5}{24} x^4+O(x^6) \end{align}$$

Hence,

$$\frac{\cos(\sin(x))-\cos(x)}{x^4}=\frac16+O(x^2)$$

from which we find the coveted limit.

score 0 · Answer 3 · answered May 13 '19 at 16:27

0

For small $x$, $\frac{\cos(\sin x)-\cos x}{\sin x-x}\sim\cos^\prime x\sim -x$, while $\frac{\sin x-x}{x^4}\sim-\frac{1}{6x}$, so the limit is $\frac16$.

answered May 13 '19 at 16:27

J.G.

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

3 Answers3