# Calculating a limit

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.

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The "$x^2$" terms in the series answers explain why you ended up using the Rule twice. –  André Nicolas Jun 21 at 12:38
@AndréNicolas, Can you explain further the connection between L'Hôpital's rule and Taylor polynomials? Moreover, how can one know a-priori what is the right order for the Taylor polynomial? (for exmple, in this case) –  AnnieOK Jun 21 at 15:42
once one has in one's mind a "library" of Taylor polynomials, it can often be clear even without formal calculation. As to connection, informally think of what happens when we apply L'Hospital's Rule to $\frac{2x^3+x^4+\cdots}{x^3+x^7+\cdots}$. –  André Nicolas Jun 21 at 17:08

$$\lim_{x\rightarrow 0} \frac{\ln(\cos x)}{x\sqrt{1+x}-x} =\lim_{x\rightarrow 0} \frac{\ln(1-\frac{x^2}{2}+o(x^2))}{x(\sqrt{1+x}-1)} =\lim_{x\rightarrow 0} \frac{-\frac{x^2}{2}+o(x^2)}{x(1+\frac{x}{2}+o(x)-1)}=$$ $$=\lim_{x\rightarrow 0} \frac{-\frac{x}{2}+o(x)}{\frac{x}{2}+o(x)}=-1$$

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Very neat. thanks! –  AnnieOK Jun 21 at 15:30
How did you get rid of the $ln$ in the top of the equation? –  Cole Johnson Jun 21 at 19:02
Using $\ln(1+y)=x+o(y)$, in this case $y=-\frac{x^2}{2}+o(x^2)$. –  Dario Jun 21 at 19:12
What is $x$ in "$x+o(y)$"? –  Elimination Aug 8 at 9:00
@Elimination It is a typo, it should be a $y$ –  Dario Aug 8 at 9:43

Using Taylor series can also prove useful, you get

$$\dfrac{-\frac{x^2}{2} + O(x^4) } { x(1+x/2)-x + O(x^3) }$$ leading straightforwardly to the result.

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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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