MHB Limit with x in both base and exponenet

[Yankel]

Dear all,

I am trying to solve the following limit:

\[\lim_{x\rightarrow 0}(e^{ax}+x)^{\frac{1}{x}}\]

where \[a\] is a constant.

I know that the limit is equal to \[e^{a+1}\] but not sure how to prove it.

Thank you.

 

[MarkFL]

I would first write:

$$L=\lim_{x\to0}\left(\left(e^{ax}+x\right)^{\frac{1}{x}}\right)$$

Next, take the natural log of both sides, and simplify to get:

$$\ln(L)=\lim_{x\to0}\left(\frac{\ln\left(e^{ax}+x\right)}{x}\right)$$

Now you have the indeterminate form 0/0 and can apply L'Hôpital's Rule. The result you seek will follow.