Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
The "$x^2$" terms in the series answers explain why you ended up using the Rule twice. – André Nicolas Jun 21 '14 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 '14 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 '14 at 17:08
up vote 5 down vote accepted

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

share|cite|improve this answer
Very neat. thanks! – AnnieOK Jun 21 '14 at 15:30
How did you get rid of the $ln$ in the top of the equation? – Cole Johnson Jun 21 '14 at 19:02
Using $\ln(1+y)=x+o(y)$, in this case $y=-\frac{x^2}{2}+o(x^2)$. – Dario Jun 21 '14 at 19:12
What is $x$ in "$x+o(y)$"? – Elimination Aug 8 '14 at 9:00
@Elimination It is a typo, it should be a $y$ – Dario Aug 8 '14 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.

share|cite|improve this answer

We need to proceed as follows $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log(\cos x)}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{\cos x - 1}\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0}1\cdot\frac{\cos x - 1}{x\sqrt{1 + x} - x}\\ &= \lim_{x \to 0} \frac{\cos x - 1}{x^{2}}\cdot\frac{x^{2}}{x\sqrt{1 + x} - x}\\ &=\lim_{x \to 0} \frac{-1}{2}\cdot\frac{\sqrt{1 + x} + 1}{1} = -1\end{aligned}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.