# A simple-looking rational limit

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!

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?? 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

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.

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

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

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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.

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