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During my homework, I've got to the following exercise:

Calculate the following limit (if exists): $$\lim \limits_{x \to 0}\left(\frac{\ln\big(\cos(x)\big)}{x^2}\right).$$

I looked in the Calculus book to find an identity that will help me resolve the expression, and what I found was:

$$\frac{1}{n}\ln(n) = \ln(\sqrt[n]n)$$

I couldn't manage to prove that Identity and thus couldn't solve the question. So what I need is:

  1. Proof to the identity,
  2. Help in solving the question.

Context: Didn't learn either L'Hopital, McLauren or Taylor Series yet. This question came just after the "Continuity" chapter.

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    $\begingroup$ Considering that you've now received three answers that use Taylor series, which you apparently haven't learned yet, please consider why adding context to your problem is important. That way, several people won't spend time writing answers that are not appropriate to your skill level. $\endgroup$ – T. Bongers Jan 12 '16 at 13:42
  • $\begingroup$ i've added De L'hospital approach .. check my answer $\endgroup$ – sirfoga Jan 12 '16 at 13:43
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    $\begingroup$ The kind of intuition you need to nurture with these problems is the ability to unpack the expressions one function at a time, and to imagine zooming in on the limit point. So $\cos(x)$, that's gonna be tending towards $1$. Now zoom into the graph of the logarithm at $x=1$. It looks like a straight line with slope one, in other words, for $x$ close to one, we basically have $\ln x = x - 1$. So the numerator's pretty much just $\cos x - 1$. And so on. $\endgroup$ – Jack M Jan 12 '16 at 17:22
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    $\begingroup$ Well the identity is just a consequence of the properties of the logarithm: you have $\ln(n) = \ln(\sqrt[n]{n} \sqrt[n]{n} \cdots \sqrt[n]{n}) = n\ln(\sqrt[n]{n})$ thus dividing both sides by $n$ we get $ln(\sqrt[n]{n}) = \frac1n \ln(n) $ $\endgroup$ – mattecapu Jan 12 '16 at 21:45
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If you can use: $$\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1$$ (which can be easily proven by considering the function $u\mapsto \ln(1+u)$, and showing that it is differentiable at $0$ with derivative equal to $1$) and $$\lim_{u\to 0}\frac{\sin u}{u} = 1$$ which also can be shown the same way $\sin^\prime 0 = \cos 0 = 1$), then you have that, for $x\in (-\frac{\pi}{2}, \frac{\pi}{2})\setminus\{0\}$:

$$ \frac{\ln\cos x}{x^2} = \frac{\ln \sqrt{1-\sin^2 x}}{x^2} = \frac{1}{2}\frac{\ln(1-\sin^2 x)}{x^2} = -\frac{1}{2}\frac{\ln(1-\sin^2 x)}{-\sin^2 x}\cdot\frac{\sin^2 x}{x^2}. $$

Using the above, since $\sin^2 x\xrightarrow[x\to0]{} 0$ , you get, as all limits exist: $$ \frac{\ln\cos x}{x^2} \xrightarrow[x\to0]{} -\frac{1}{2}\left(\lim_{u\to 0}\frac{\ln (1+u)}{u}\right)\cdot \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2 = -\frac{1}{2} $$

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  • $\begingroup$ Note that the first step, showing that $\lim_{u\to 0}\frac{\ln (1+u)}{u} = 1$, is equivalent to a Taylor expansion of $\ln(1+u)$ to order $1$. But it only uses elementary calculus, and does not rely on even knowing what a Taylor expansion is. $\endgroup$ – Clement C. Jan 12 '16 at 14:18
  • $\begingroup$ Fantastic answer! In my opinion, this is more elegant than using taylor series and whatnot. $\endgroup$ – Airdish Jan 12 '16 at 14:26
  • $\begingroup$ Thank you. I know derivatives so I can understand your answer. But I didn't learn in yet so I can't use it in my homework $\endgroup$ – CodeNinja Jan 12 '16 at 14:29
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Use Taylor's formula at order $2$ and equivalents:

  • $\cos x=1-\dfrac{x^2}2+o(x^2)$
  • $\ln(1-u)\sim_0 -u$, hence $\ln(\cos x)\sim_0 -\dfrac{x^2}2+o(x^2)\sim_0 -\dfrac{x^2}2$.

From this, we deduce $$\frac{\ln\cos x}{x^2}\sim_0 -\frac{x^2}{2x^2}=-\frac12.$$

Without Taylor's formula:

You can prove directly that $\;\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{\sin^2 x }{x^2(1+ \cos x)}=\frac12$, so that $$\frac{1-\cos x}{x^2}\sim_0\frac12,\quad\text{whence}\quad \cos x\sim_0 1 - \dfrac{x^2}2.$$

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  • $\begingroup$ I can't use it because I didn't learn it yet $\endgroup$ – CodeNinja Jan 12 '16 at 13:39
  • $\begingroup$ Which is first: L'Hospital rule or Taylor expsansion? $\endgroup$ – kmitov Jan 12 '16 at 13:39
  • $\begingroup$ You didn't learn what? $\endgroup$ – Bernard Jan 12 '16 at 13:40
  • $\begingroup$ They Taylor's formula. This question came right after the Continuity chapter. $\endgroup$ – CodeNinja Jan 12 '16 at 13:51
  • $\begingroup$ Then, in any case keep that in mind for when you will -- it's never lost. $\endgroup$ – Clement C. Jan 12 '16 at 13:52
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There are many ways to solve this question:

  • have you heard about Taylor series? By applying McLaurin rules you get $$\lim \limits_{x \to 0}\frac{\ln(\cos(x))}{x^2} = \lim \limits_{x \to 0}\frac{\ln \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^4)\right)}{x^2}\\=\lim \limits_{x \to 0}\frac{-\frac{x^2}{2!} + O(x^2)}{x^2} = \lim \limits_{x \to 0}\frac{x^2 \cdot \left( -\frac{1}{2} + \frac{O(x^2)}{x^2}\right)}{x^2} = \lim \limits_{x \to 0}-\frac{1}{2} + \frac{O(x^2)}{x^2} = -\frac{1}{2}$$
  • another approach is using L'Hospital's rule: differentiate both numerator and denominator (given you checkes the hypothesis) and you get $$\lim \limits_{x \to 0}\frac{\ln(\cos(x))}{x^2} = \lim \limits_{x \to 0}\frac{\frac{1}{\cos x} \cdot -\sin x}{2x}$$ and since $ \lim \limits_{x \to 0}\frac{\sin x}{x} = 1$ you can simplify to get $$$\lim \limits_{x \to 0}\frac{\ln(\cos(x))}{x^2} = \lim \limits_{x \to 0}\frac{\frac{1}{\cos x} \cdot -\sin x}{2x} = \lim \limits_{x \to 0}-\frac{1}{2 \cos x} = -\frac{1}{2}$$
  • classic approach is to check the limit by its definition: let $f(x) = \frac{\ln(\cos(x))}{x^2}$, what you want to check is that for every number $\epsilon$ there is some number $\delta$ such that $$|f(x) - \left( -\frac{1}{2} \right) | < \epsilon$$ whenever $| x - 0| < \delta$
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  • $\begingroup$ But this is another question. $\endgroup$ – kmitov Jan 12 '16 at 13:35
  • $\begingroup$ Yu can apply L'Hospital rule. $\endgroup$ – kmitov Jan 12 '16 at 13:35
  • $\begingroup$ what you can use? $\endgroup$ – kmitov Jan 12 '16 at 13:36
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    $\begingroup$ Can you elaborate about the last "classic" approach? $\endgroup$ – CodeNinja Jan 12 '16 at 14:00
  • $\begingroup$ Classic approach is wrong... one that's the derivative, two it's misleading to just restate the question and call that the "classic approach" next to two actual approaches. $\endgroup$ – djechlin Jan 12 '16 at 21:14
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Use the well-known Taylor series $$\cos(x)= 1-\frac{1}{2}x^2+\frac{1}{24}x^4+O(x^6)$$ $$\ln(1+x)= x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5+O(x^6)$$ to get $$\ln(\cos(x))=-\frac{1}{2}x^2-\frac{1}{12}x^4+O(x^6)$$ Can you continue?

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    $\begingroup$ FWIW: no need to go as far as order $4$ (which looks a tad scary at first). $\endgroup$ – Clement C. Jan 12 '16 at 13:52

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