I am trying to use L'Hopital's rule to evaluate the following limit but I am not sure I am doing it correctly.
$\lim_{x \to 0} \left(e^x+x\right)^{\frac{1}{x}}.$
To use L'Hopitals rule I have been told that the limit has to evaluate to certain in-determinants before it can be applied, in the case above we would get $1^{\infty}$ so it can be used.
Here are my steps to evaluate the limit:
$\lim_{x \to 0} \left(e^x+x\right)^{\frac{1}{x}}.$
$\lim_{x \to 0} \ln\left(\left(e^x+x\right)^{\frac{1}{x}}\right) = \lim_{x \to 0} \frac{1}{x} \cdot \ln\left(e^x+x\right)$ $\lim_{x \to 0} \frac{\ln\left(e^x+x\right)}{x} = \lim_{x \to 0} \frac{1}{x^2+xe^x}$ $\lim_{x \to 0} \frac{0}{2x+e^x+xe^x} = \frac{0}{1}=0$
As I took the natural log at the start the evaluated limit would be $e^0=1$.
So I have two questions, firstly is my method correct? I am particularly unsure about what to do when (for example in the 4th line) I get a 0 as my numerator, is $\frac{0}{1}$ considered to be an in determinant? Secondly if the method is correct, is my answer correct? Is there a way to prove what I have done is the limit evaluated?
I am not sure on the definition of an in-determinant as in my lecture notes I have that I can apply L'Hopital's rule on problems with $1^{\infty}$ but I am not sure if this is an in-determinant in the same way that $\frac{0}{0}$ is considered to be undefined. I am not sure my last point is entirely clear so if it is not please let me know and I will try to clarify further.