`lim_(xto0)((e^(x)-1)/x)^(1//x)`

Options

A. e
B. `e^(2)`
C. `3sqrt(e)`
D. `sqrt(e)`

Answer

To solve the limit \( \lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} \), we will follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ L = \lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} \]
This limit is in the form of \( f(x)^{g(x)} \) where \( f(x) = \frac{e^x - 1}{x} \) and \( g(x) = \frac{1}{x} \).

### Step 2: Apply the Exponential Limit Property
We can rewrite this limit using the property of limits:
\[ L = e^{\lim_{x \to 0} g(x) \cdot (f(x) - 1)} \]
Thus, we need to find:
\[ \lim_{x \to 0} \frac{1}{x} \left( \frac{e^x - 1}{x} - 1 \right) \]

### Step 3: Simplify the Expression
We simplify \( \frac{e^x - 1}{x} - 1 \):
\[ \frac{e^x - 1 - x}{x} \]
Now we need to evaluate:
\[ \lim_{x \to 0} \frac{e^x - 1 - x}{x^2} \]

### Step 4: Apply L'Hôpital's Rule
Since the limit is in the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
Differentiate the numerator and denominator:
- The derivative of \( e^x - 1 - x \) is \( e^x - 1 \).
- The derivative of \( x^2 \) is \( 2x \).

Thus, we have:
\[ \lim_{x \to 0} \frac{e^x - 1}{2x} \]

### Step 5: Apply L'Hôpital's Rule Again
This limit is still in the form \( \frac{0}{0} \), so we apply L'Hôpital's Rule again:
- The derivative of \( e^x - 1 \) is \( e^x \).
- The derivative of \( 2x \) is \( 2 \).

Now we have:
\[ \lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2} \]

### Step 6: Substitute Back into the Exponential
Now substituting back into our limit:
\[ L = e^{\frac{1}{2}} = \sqrt{e} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \left( \frac{e^x - 1}{x} \right)^{\frac{1}{x}} = \sqrt{e} \]