I have a function $$f(x)=\frac{2(x-\log(x+1))}{x^2}$$ which is not defined at $x=0$, but I can use L'Hopistal rule two times to get the limit as $x\rightarrow 0$:
$$\lim_{x\rightarrow 0} \frac{2(x-\log(x+1))}{x^2} =\lim_{x\rightarrow 0} \frac{1-\frac{1}{x+1}}{x} = \lim_{x\rightarrow 0} \frac{\frac{1}{(x+1)^2}}{1}=1$$
But then I tried to plot the function on Desmos, and got this graph
Clearly this graph doesn't approach $1$ as $x\rightarrow 0$, so I thought I made a mistake, but when I put the same function in WolframAlpha, I got this:
Does anybody know what's going on here?
The trick lies in the definition of $\log$ and $\ln$. The limit you have derived is true for $\log_e$ which is $\ln$. But $\log$ (that is used in Desmos) means $\log_{10}$. Now if you recalculate your limit using $\log_{10}$ you will get the same result as shown in Desmos. On the other hand Wolfram probably used $\log$ with the base $e$. You can verify it by checking the value of $\log e$ in Wolfram. It will give $1$. But in Desmos $\log e$ should be $0.434294481903$. Try applying $\ln$.