<s>[INST] Find $\lim_{x \to 0} \left( \frac{\ln (\cos x)}{x\sqrt {1 + x} - x} \right)$ efficiently 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? 
$$\ln\left(1- {x^2 \over 2}\right) \approx -{x^2 \over 2}$$
 [/INST] 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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