Limit with x in both base and exponenet

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SUMMARY

The limit of the expression \(\lim_{x\rightarrow 0}(e^{ax}+x)^{\frac{1}{x}}\) evaluates to \(e^{a+1}\). To prove this, first define \(L=\lim_{x\to0}\left(\left(e^{ax}+x\right)^{\frac{1}{x}}\right)\). By taking the natural logarithm of both sides, the limit simplifies to \(\ln(L)=\lim_{x\to0}\left(\frac{\ln\left(e^{ax}+x\right)}{x}\right)\), which presents an indeterminate form of 0/0, allowing the application of L'Hôpital's Rule to derive the final result.

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Yankel
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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.
 
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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.
 

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