<s>[INST] Evaluate the limit $\lim_{x\rightarrow 0}\left[ \frac{\ln(\cos x)}{x\sqrt{1+x}-x} \right]$ Evaluate the limit:
$$\lim_{x\rightarrow 0}\left[ \frac{\ln(\cos x)}{x\sqrt{1+x}-x} \right]$$
I actually was able to find the limit is $-1$ after applying L'Hôpital's rule twice.
I wonder if that was the intention of this exercise or there's an "easier" way.
Thanks.
 [/INST] We need to proceed as follows $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log(\cos x)}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{\cos x - 1}\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0}1\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\
&= \lim_{x \to 0} \frac{\cos x - 1}{x^{2}}\cdot\frac{x^{2}}{x\sqrt{1 + x} - x}\\
&=\lim_{x \to 0} \frac{-1}{2}\cdot\frac{\sqrt{1 + x} + 1}{1} = -1\end{aligned}$$
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