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I need to evaluate:
$$\lim_{x \to 0} \left( \frac{\ln (\cos x)}{x\sqrt {1 + x} - x} \right)$$

Now, it looked to me like a classic L'Hôpital's rule case. Indeed, I used it (twice), but then things became messy and complicated.

Am I missing the point of this exercise? I mean, there must be a "nicer" way. Or should I stick with this road?


Regarding Yiorgos's answer: Why is the following true? $$\ln\left(1- {x^2 \over 2}\right) \approx -{x^2 \over 2}$$

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Are you sure you aren't missing an absolute value or something? – Tim Seguine Jan 25 '14 at 22:26
Problem: the $\log\cos$ is $-\infty$ in the roots of $\cos$. – Martín-Blas Pérez Pinilla Jan 25 '14 at 22:26
@Martín-BlasPérezPinilla and undefined for half of the real line(alternating intervals of length pi) – Tim Seguine Jan 25 '14 at 22:27
I changed \mathop{lim}\limits_{x\to0} to \lim_{x\to0}. That is standard. Also, you only need as many curly braces as you need. Writing things like {{{x}^{2}}} where x^2 suffices is just clutter. There were some things along those lines and I cleaned them up. – Michael Hardy Jan 25 '14 at 22:34
In reference to the question in your edit: that is the first term of its Taylor Series. – Tim Seguine Jan 25 '14 at 23:17
up vote 4 down vote accepted


I. $\ln \cos x\approx \ln \Big(1-\frac{x^2}{2}\Big)\approx -\frac{x^2}{2}$

II. $$x\sqrt{x+1}-x=\frac{x^2(x+1)-x}{x\sqrt{1+x}+x}=\frac{x(x+1)-1}{\sqrt{1+x}+1}\approx -\frac{1}{2}.$$

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Oh I think I get it. It's the Taylor expansion for $cos(x)$. But why are you allowed to do that? (developing till order "2" if I have not mistaken). – SuperStamp Jan 25 '14 at 22:45
Taylor expansion is how l'hopital's rule works. All you are doing is comping the leading order terms of the taylor polynomial. This works because each term in an expansion is successive derivatives of the original function. – Richard P Jan 25 '14 at 23:05
@SuperStamp, expand the Taylor series for $\log(x+1)$ and you get $x - x^2/2 + x^3/3 ...$. Hence, $\log(1-x) ~ -x$. – Chris K Jan 25 '14 at 23:06
Oh right. Thanks @ChrisK – SuperStamp Jan 25 '14 at 23:07
@RichardP, Oh thanks. Good to know the connection between the two. – SuperStamp Jan 26 '14 at 2:12

Multiply the numerator and denominator by $(\sqrt{1-x}+1).$ Then use L'Hospital's Rule.

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Using that (see

$$ \lim_{x\to 0}{\ln\cos x\over\cos x - 1}=1\qquad{\rm and} \qquad\lim_{x\to 0}{\cos x - 1\over -x^2/2}=1, $$ we have: $$ \lim_{x\to 0}\frac{\ln(\cos x)}{x\sqrt{1+x}-x}= \lim_{x\to 0}\frac{\cos x - 1}{x\sqrt{1+x}-x}= \lim_{x\to 0}\frac{-x^2/2}{x\sqrt{1+x}-x}=\cdots $$

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What do you mean $\cos x \approx \cos x - 1$. I don't get it. Can you expand your answer please? – SuperStamp Jan 25 '14 at 22:48

By l'Hopital: $\lim_{x\rightarrow 0}(.)=\lim_{x\rightarrow 0}\frac{-tan(x)}{\frac{x}{2\sqrt{1+x}}+\sqrt{1+x}-1}$

With a second l'Hopital:

$=\lim_{x\rightarrow 0}(.)\frac{-sec^2(x)}{\frac{3x+4}{4(x+1)^{3/2}}}=-1$

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How did you evaluate this as $-2$? – SuperStamp Jan 25 '14 at 22:41
Her answer is wrong; you should get $-1$. – Chris K Jan 25 '14 at 22:45
Sorry, you need to take l'Hopital twice. I just fixed it. – Alt Jan 25 '14 at 23:00

One trick you can use for tricky limits like this is to taylor expand the whole function. In this case it is pretty nasty, but you wind up with the result of: $$\lim_{x\rightarrow0}(-1 - \frac{x}{4} + O(x^2))$$ Since this expansion is equal to the original function, it is immediately clear that the limit as it approaches 0 is -1.

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By direct computation:

$$\lim_{x \rightarrow 0^{+}} \frac{\log(\cos x)}{x(\sqrt{x+1}-1)} = \\ \lim \frac{-\tan x}{\frac{x}{2\sqrt{x+1}}+\sqrt{x+1}-1} = \\ \lim \frac{-\sec^2 x}{\frac{\sqrt{x+1}-x/(2\sqrt{x+1})}{2(x+1)}+\frac{1}{2\sqrt{x+1}}} = \\ \frac{-1}{1/2+1/2} = -1$$

But surely there is a nicer or more elegant method?

I note that only the one-sided limit exists unless of course you are evaluating:

$$lim_{x \rightarrow 0}\; \Re [\frac{\log(\cos x)}{x(\sqrt{x+1}-1)}]$$

However, I digress.

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Maybe I should've stick with that. Not so bad after all. But I did learn from the answers here, so it was worth. Thanks! – SuperStamp Jan 25 '14 at 22:57
What would happen when $x\rightarrow 0^-$? – SuperStamp Jan 25 '14 at 23:06
You would have the same 'real' part; however, you inherit an imaginary part. – Chris K Jan 25 '14 at 23:07

$$x\sqrt{1 + x} - x = x(\sqrt{1 + x} - 1) = x\left({1\over 1 + \sqrt{1 + x}}\right)\sim \sqrt{x} $$ as $x\to\infty$. That is the nice part. The numerator is an irremediable problem.

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I appreciate your help, tough a mistake was made. the limit is $x\rightarrow 0$. – SuperStamp Jan 25 '14 at 22:29

Elaborating on Martin's answer, the calculation of limit is as follows:

$\displaystyle \begin{aligned}L &= \lim_{x \to 0}\frac{\log(\cos x)}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0}\frac{\log(\cos x)}{\cos x - 1}\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0}1\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0}\frac{\cos x - 1}{x\left(\sqrt{1 + x} - 1\right)}\\ &= \lim_{x \to 0}\frac{\cos x - 1}{x\left(\sqrt{1 + x} - 1\right)}\cdot\frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}\\ &= \lim_{x \to 0}\frac{\cos x - 1}{x^{2}}\cdot\left(\sqrt{1 + x} + 1\right)\\ &= -\frac{1}{2}\cdot 2 = -1\end{aligned}$

Note that if $y = \cos x - 1$ then $y \to 0$ as $x \to 0$ and hence $\dfrac{\log(1 + y)}{y} = \dfrac{\log \cos x}{\cos x - 1}$ tends to $1$. The other limit $\dfrac{\cos x - 1}{x^{2}}$ is easily done by using $\cos x - 1 = -2\sin^{2}(x/2)$ and then using the fundamental limit $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$.

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