I have to prove $\lim_{x\to0}(1+x)^{1/x}=e$. Now I would like to perform the following operations:
$$\lim_{x\to0}(1+x)^{1/x}=e$$$$\lim_{x\to0}\ln(1+x)^{1/x}=1$$$$\lim_{x\to0}\frac{\ln(1+x)}x=1$$ And then using L'Hôpital's rule $$\lim_{x\to0}\frac{1}{1+x}=1$$
But the first step doesn't seem kosher, as I'm taking the logarithm "over" the limit, is there some way to justify that?
Yes, it is kosher, because every value of $(1+x)^{1/x}$ for $x$ a positive real number is a positive real number, and $\ln$ is defined and continuous on positive real numbers. As it happens the limit of $(1+x)^{1/x}$ as $x\searrow 0$ is still a positive real number; if the limit of some positive function were $0$ as $x\searrow 0$, you would still get it right, because the logarithm of the function would tend to $-\infty$.
Yes but you don't have to. Instead you can expand log in Taylor series:$x + O(x^2)$ and, given than $t =e^{\log t}$ the result comes out.
