# Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series

Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.

$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$

Any tips will be helpful.

Sorry, but I don't want to use the Taylor series as well.

• that diverges to $\infty$ surely? Jan 4, 2020 at 12:37
• No, it doesn't, it converges to $-\frac{1}{6}$ Jan 4, 2020 at 12:38
• No, it really diverges to $\infty$ as the numerator is $\sim x^2/2$. Jan 4, 2020 at 12:40
• @Anonymous You have $\frac{1}{x^2} + \frac{\log(\cos x)}{x^4}$. How could that converge to a constant? Jan 4, 2020 at 12:42
• Sorry I forgot a $2$ Jan 4, 2020 at 12:47

Result 1: $$\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} = \frac{1}{3}$$

Proof. Note that $$\sin x = x - \frac{x^3}{3!} + o(x^5).$$

Thus, $$\sin^2x = x^2 - 2x\frac{x^3}{3!} + o(x^5).$$

This, gives $$x^2 - \sin^2 x = \frac{x^4}{3} + o(x^5),$$ and the result follows.

Result 2: $$\displaystyle\lim_{x\to0} \dfrac{\sin^4x}{x^4} = 1$$

Proof. Follows trivially from $$\displaystyle \lim_{x\to0} \frac{\sin x}{x} = 1.$$

Result 3: $$\ln(1 - x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} + o(x^4).$$ (Expansion is valid near $$0$$)

Proof. Standard result. This is the Taylor expansion of $$\ln(1-x)$$ near $$0$$.

Solution.

$$\displaystyle\lim_{x\to0}\dfrac{x^2 + 2\ln(\cos x)}{x^4}$$

$$=\displaystyle\lim_{x\to0}\dfrac{x^2 + \ln(\cos^2 x)}{x^4}$$

$$=\displaystyle\lim_{x\to0}\dfrac{x^2 + \ln(1 - \sin^2 x)}{x^4}$$

$$=\displaystyle\lim_{x\to0}\dfrac{x^2 + (-\sin^2x - \frac{\sin^4x}{2} + o(x^6))}{x^4}$$

$$=\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} - \dfrac{1}{2}\displaystyle\lim_{x\to0}\dfrac{\sin^4x}{x^4} + 0$$

$$=\dfrac{1}{3} - \dfrac{1}{2}$$

$$=\boxed{-\dfrac{1}{6}}$$

• Interesting. Wolfram says $0$ but Symbolab says $-\dfrac{1}{6}$. Hmm either Wolfram or my input were incorrect... Jan 4, 2020 at 13:22

$$F=\lim_{x\to0}\dfrac{x^2+\ln(1-\sin^2x)}{x^4}$$

$$=\lim\dfrac{x^2-\sin^2x-(\sin^2x)^2/2+O(x^6)}{x^4}$$

$$=-\dfrac12+\lim\dfrac{x-\sin x}{x^3}\dfrac{x+\sin x}x$$

Hint:

Use Taylor expansion at order $$4$$: as $$\cos x=1-\frac{x^2}2+\frac{x^4}{24}+o(x^4),$$ setting $$u=-\dfrac{x^2}2+\dfrac{x^4}{24}+o(x^4)$$, we have to expand $$\ln (1+u)$$ at order $$2$$ in $$u$$ and truncate the result at order $$4$$ (in $$x$$): \begin{align} \ln(\cos x)&=\ln(1+u)=u-\frac{u^2}2+o(u^2)=-\dfrac{x^2}2+\dfrac{x^4}{24}-\frac12\biggl(-\dfrac{x^2}2+\dfrac{x^4}{24}\biggr)^2+o(x^4)\\ &=-\dfrac{x^2}2+\dfrac{x^4}{24}-\dfrac{x^4}{8}+o(x^4)=-\dfrac{x^2}2-\dfrac{x^4}{12}+o(x^4) \end{align} so that the numerator is $$x^2+2\ln(\cos x) =-\dfrac{x^4}{6}+o(x^4)\sim_0 -\dfrac{x^4}{6}.$$

• You forgot to halve $u^2$ when you worked to $O(x^4)$, so your coefficient is wrong. It should be $\frac{1}{24}-\frac18=-\frac{1}{12}$, giving the OP's stated answer of $-\frac16$.
– J.G.
Jan 4, 2020 at 13:19
• Oh! yes. I shouldn't type directly on screen. 'Tis fixed. Thanks for pointing it (and a happy newyear!). Jan 4, 2020 at 13:27