Basically the title.
The solution in the book I am reading says to look at the function $f(x) = \ln(1+dx)$. Clearly, $f'(0) = \lim_{x \to 0} \frac{\ln(1+dx)}{x} = d$. I follow so far. But then, the author says "Applying the continuous function $e^x$" yields $\lim_{x \to 0} (1+dx)^{\frac{1}{x}} = e^d$.
I got completely lost here. I have no idea how the author applied $e^x$ on the LHS to get the new limit. I understand once I have this new limit what I have to do to prove the rest, but I am not sure how the application works. I may just not be remembering some log rules, but some guidance would be very appreciated.
Thank you!
As the exponential function is continuous, the author uses $$ \lim_{x\to0}g(x)=a\implies \lim_{x\to0}e^{g(x)}=e^d.$$ Here, $g(x)=\frac{\ln(1+dx)}x$, so $$e^{g(x)}=e^{\frac{\ln(1+dx)}x}=\left(e^{\ln(1+dx)}\right)^{1/x} =(1+dx)^{1/x}$$