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.
Best Answer
L’Hospital’s rule applies only to limits of the form $0/0$ and $\infty/\infty$. However, there is a standard trick for converting the indeterminate form $1^\infty$ to one of these forms so that l’Hospital’s rule can be applied, and that’s essentially what you’re doing here. Let me write it up in a way that makes a little clearer exactly what is going on.
Let $L=\displaystyle\lim_{x\to 0}\left(e^x+x\right)^{1/x}$. Then $$\ln L=\ln\lim_{x\to 0}\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}\;,$$ since $\ln$ is a continuous function. Thus,
$$\ln L=\lim_{x\to 0}\ln\left(e^x+x\right)^{1/x}=\lim_{x\to 0}\frac1x\ln(e^x+x)=\lim_{x\to 0}\frac{\ln(e^x+x)}x\;.$$
This last limit is a $\frac00$ indeterminate form, so l’Hospital’s rule applies:
$$\ln L=\lim_{x\to 0}\frac{\ln(e^x+x)}x=\lim_{x\to 0}\frac{\frac{e^x+1}{e^x+x}}1=\lim_{x\to 0}\frac{e^x+1}{e^x+x}=\frac21=2\;.$$
Finally, then $L=e^{\ln L}=e^2$.
You went astray when you wrote $$\lim_{x \to 0} \frac{\ln\left(e^x+x\right)}{x} = \lim_{x \to 0} \frac{1}{x^2+xe^x}\;:$$ you did not differentiate the numerator correctly, and you did not differentiate the denominator at all.
If $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=1$, then $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac01=0$; there is nothing indeterminate here.