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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.

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

3 Answers 3

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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.

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

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

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

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

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

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

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

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  • $\begingroup$ Interesting. Wolfram says $0$ but Symbolab says $-\dfrac{1}{6}$. Hmm either Wolfram or my input were incorrect... $\endgroup$
    – Déjà vu
    Jan 4, 2020 at 13:22
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Using http://mathworld.wolfram.com/SeriesExpansion.html

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

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

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

Use Are all limits solvable without L'Hôpital Rule or Series Expansion

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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 $$x^2+2\ln(\cos x) =-\dfrac{x^4}{6}+o(x^4)\sim_0 -\dfrac{x^4}{6}.$$

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  • $\begingroup$ 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 $-\frac16$. $\endgroup$
    – J.G.
    Jan 4, 2020 at 13:19
  • $\begingroup$ Oh! yes. I shouldn't type directly on screen. 'Tis fixed. Thanks for pointing it (and a happy newyear!). $\endgroup$
    – Bernard
    Jan 4, 2020 at 13:27

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