Find $\lim_{x \to 0} \left( \frac{\ln (\cos x)}{x\sqrt {1 + x} - x} \right)$ efficiently

Question

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?

EDIT:

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

Answer 1

Hints.

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

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

Answer 2

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

Answer 3

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:

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

Answer 4

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

Answer 5

Using that (see http://www.math24.net/infinitesimals.html)

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

Answer 6

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.

Answer 7

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.

Answer 8

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 $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$.