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To evaluate the limit:

\[
\lim_{x \to 0} \left( a^x + x \right)^{1/x}
\]

we can start by taking the natural logarithm to simplify the expression. Let:

\[
L = \lim_{x \to 0} \left( a^x + x \right)^{1/x}
\]

Taking the natural logarithm of both sides, we get:

\[
\ln L = \lim_{x \to 0} \frac{\ln(a^x + x)}{x}
\]

We can apply L'Hôpital's rule to evaluate this limit because it is of the indeterminate form \(\frac{0}{0}\) as \(x \to 0\).

Let's find the derivative of the numerator and the denominator. The derivative of \(\ln(a^x + x)\) with respect to \(x\) is:

\[
\frac{d}{dx} \ln(a^x + x) = \frac{a^x \ln a + 1}{a^x + x}
\]

The derivative of the denominator \(x\) with respect to \(x\) is 1.

Now applying L'Hôpital's rule:

\[
\ln L = \lim_{x \to 0} \frac{a^x \ln a + 1}{a^x + x}
\]

As \(x \to 0\), \(a^x \to 1\), so we have:

\[
\ln L = \frac{1 \cdot \ln a + 1}{1 + 0} = \ln a + 1
\]

Therefore,

\[
L = e^{\ln a + 1} = a \cdot e
\]

So the limit is:

\[
\lim_{x \to 0} \left( a^x + x \right)^{1/x} = a \cdot e
\]

Would you like more details or have any questions? Here are some related questions to consider:

1. What is L'Hôpital's rule and how is it used in limit evaluation?
2. How does taking the natural logarithm help in simplifying limits involving exponents?
3. Can you provide another example of a limit that requires L'Hôpital's rule for evaluation?
4. What are indeterminate forms and why are they important in calculus?
5. How does the behavior of \(a^x\) change as \(x \to 0\) for different values of \(a\)?

**Tip:** When evaluating complex limits, taking the natural logarithm can often simplify the problem by turning exponentiation into multiplication, which can then be managed using L'Hôpital's rule or other techniques.

Evaluate the limit: lim(x → 0) (a^x + x)^(1/x)

Learn how to evaluate the limit lim(x → 0) (a^x + x)^(1/x) using L'Hopital's Rule. This problem involves calculus concepts such as limits, exponential functions, and logarithmic differentiation, suitable for students in advanced high school or college-level calculus.

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