A simple-looking rational limit

Question

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!

Answer 1

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.

Answer 2

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

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

Answer 3

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.