Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me compute: $$ \lim_{z\to 0}\frac{\sqrt{2(z-\log(1+z))}}{z} $$ I know the answer is 1 because I plugged it into Mathematica. Attempts with L'Hopital's Rule didn't work. This a step in an exercise for my self-study project. Thanks!

share|improve this question
1  
?? According to the WolframAlpha limit does not exist: wolframalpha.com/input/… –  georg Aug 22 at 6:40
    
Yes @georg you are right because for negative $z$'s limit can't be $1$. –  user121270 Aug 22 at 6:41
    
You guys are right, I should have added sign(z) to the numerator. –  Kim Jong Un Aug 23 at 5:06

3 Answers 3

up vote 3 down vote accepted

Using Taylor series $$\log (1+z)\sim_0 z-\frac{z^2}{2}$$ we get

$$\frac{\sqrt{2(z-\log(1+z))}}{z}\sim_0\frac{|z|}{z}\sim\left\{\begin{array}[cc]\\1\;&\text{at}\; 0^+\\-1\;&\text{at}\; 0^-\end{array}\right.$$ so the limit doesn't exist.

share|improve this answer

$$\log(1+z)=z-\frac{z^2}2+\frac{z^3}3-\cdots$$

So, $$z-\log(1+z)=\frac{z^2}2-\frac{z^3}3+\cdots$$

share|improve this answer
    
Thanks Lab. If you have time, could you please write out a formal justification for switching $\lim$ and $\sum$ in $\lim_{z\to 0}\sum_{n=2}^\infty\frac{z^{n-2}(-1)^n}{n}$? I am guessing it likely involves uniform convergence but my calculus is so rusty at the moment, I can't write a neat justification myself. –  Kim Jong Un Aug 22 at 7:04

Apply L'Hopital's rule to the square of the function: $$\lim_{z\to 0}\frac{2(z-\log(1+z))}{z^2}=\lim_{z\to 0}\frac{2(1-\frac{1}{1+z})}{2z}=\cdots=1\ .$$ See if you can fill in the working and then finish your problem for $z\to0^+$.

In fact, if you look carefully at the limit as $z\to0^-$ you will find it is different. So the (proper, "two-sided") limit does not exist.

share|improve this answer

Your Answer

 
discard

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.