this is my 3rd question of the day but so far you guys have been helping me a lot, so I hope it isn't too much to ask about 1 more, so i have this limit
$$\lim_{x\to0}{\frac{\ln(2-\cos(x))}{\cosh(x)-1}}$$
and i've calculated(I am not allowed to use L'hospital's) as follows $$\lim_{x\to0}{\frac{\ln(2-\cos(x))}{\cosh(x)-1}}=\lim_{x\to0}{\frac{\ln((-\cos(x)+1)+1)}{-\cos(x)+1}\frac{(-\cos(x)+1)}{\cosh(x)-1}\to\lim_{x\to0}{\frac{-\cos(x)+1}{\cosh(x)-1}=\frac{\ln(-\cos(x)+1)}{-2\ln(2)-1}}}\to\frac{\ln(0)}{-2\ln(2)-1}\to\frac{\infty}{-2\ln(2)-1}\to\infty$$ by using $\lim_{x\to0}\frac{\ln(x_n+1)}{x_n}=1$ and the definition of the $\cosh(x)$ function
and then I tried this way after knowing the limit was 1 $$\frac{\ln(2-\cos(x))}{\cosh(x)-1}=\frac{2-\cos(x)}{\exp(\frac{e^x+e^{-x}}{2}-1)}\to_{\lim_{x\to0}}\frac{1}{1}$$
I would like to understand why is the first method incorrect or even if the second one is also correct, where can I see that the logic "falls off"?
As always thank you very much in advance
The second approach is wrong because $$\frac{\ln(2-\cos(x))}{\cosh(x)-1}=\frac{2-\cos(x)}{\exp\left(\frac{e^x+e^{-x}}{2}-1\right)}$$ is wrong. Are you trying to use the incorrect property $e^{a/b}=e^a/e^b$?
As for the first one, the line $$\lim_{x\to0}{\frac{\ln((-\cos(x)+1)+1)}{-\cos(x)+1}\frac{(-\cos(x)+1)}{\cosh(x)-1}\to\lim_{x\to0}{\frac{-\cos(x)+1}{\cosh(x)-1}=\frac{\ln(-\cos(x)+1)}{-2\ln(2)-1}}}$$ doesn't make sense to me. First, a limit doesn't approach another limit. A limit, if it exists, is a number and a limit can equal another limit. So that arrow shouldn't be there. Next, I can't even tell how the second limit became equal to what you wrote.
Here's how this method can work: $$\frac{\ln(2-\cos(x))}{\cosh(x)-1}=\frac{\ln(1+(1-\cos(x))}{1-\cos(x)}\frac{1-\cos(x)}{x^2}\frac{x^2}{\cosh(x)-1} $$ Then compute each of the limits separately $$\lim_{x\to 0}\frac{\ln(1+(1-\cos(x))}{1-\cos(x)}\qquad\lim_{x\to 0}\frac{1-\cos(x)}{x^2}\qquad\lim_{x\to 0}\frac{x^2}{\cosh(x)-1} $$ to find the required limit $$\begin{align}\lim_{x\to 0}\frac{\ln(2-\cos(x))}{\cosh(x)-1}&=\lim_{x\to 0}\frac{\ln(1+(1-\cos(x))}{1-\cos(x)}\cdot\lim_{x\to 0}\frac{1-\cos(x)}{x^2}\cdot\lim_{x\to 0}\frac{x^2}{\cosh(x)-1}\\&=L_1\cdot L_2\cdot L_3\end{align}$$