State the indeterminate form of the limit of (e^x + x)^(1/x) as x approaches 0, then use L Hopital rule to evaluate it.

Step-by-step solution

1. 1

   Identify the indeterminate form. As $x\to 0$, the base $e^x + x \to 1$ and the exponent $\tfrac{1}{x} \to \pm\infty$. So the form is $1^{\infty}$ — indeterminate, because a base slightly above $1$ raised to a huge power can approach anything. It is not simply $1$.

2. 2

   Take logarithms to convert the power into a quotient. Let $L$ be the limit and set $y = (e^x+x)^{1/x}$:

   $\ln y = \frac{\ln(e^{x}+x)}{x}$

   As $x\to 0$ the numerator $\to \ln 1 = 0$ and the denominator $\to 0$, so this is now $\tfrac{0}{0}$ and L'Hopital's Rule applies.

3. 3

   Differentiate top and bottom. For the numerator use the chain rule:

   $\frac{d}{dx}\ln(e^x+x) = \frac{e^{x}+1}{e^{x}+x}, \qquad \frac{d}{dx}x = 1$

4. 4

   Evaluate the new quotient at $x = 0$. It is no longer indeterminate:

   $\lim_{x\to 0}\frac{e^{x}+1}{e^{x}+x} = \frac{1+1}{1+0} = \frac{2}{1} = 2$

   So $\ln L = 2$.

5. 5

   Undo the logarithm. This final exponentiation is the step most often forgotten, leaving the answer as $2$:

   $L = e^{2} \approx 7.389056$

6. 6

   Verify numerically from both sides. At $x = 0.001$: $e^{0.001}+0.001 = 1.0020010$, and $1.0020010^{1000} = 7.396$. At $x = -0.001$: $e^{-0.001}-0.001 = 0.9980005$, and $0.9980005^{-1000} = 7.382$. Both bracket $e^2 = 7.38906$ ✓, and a symbolic limit returns exactly $e^2$.