`lim_(lamda to 0)(int_(0)^(1) (1+x)^(lambda ) dx)^(1//lambda)` Is equal to

Options

A. (a) `2 In 2`
B. (b) `(4)/(e)`
C. (c) `In(4)/(e)`
D. (d) 4

Answer

To solve the limit

\[ \lim_{\lambda \to 0} \left( \int_{0}^{1} (1+x)^{\lambda} \, dx \right)^{\frac{1}{\lambda}}, \]

we will follow these steps:

### Step 1: Evaluate the Integral
We start by evaluating the integral

\[ I(\lambda) = \int_{0}^{1} (1+x)^{\lambda} \, dx. \]

Using the formula for integration, we have:

\[ I(\lambda) = \left[ \frac{(1+x)^{\lambda + 1}}{\lambda + 1} \right]_{0}^{1}. \]

Calculating the limits, we find:

\[ I(\lambda) = \frac{(1+1)^{\lambda + 1} - (1+0)^{\lambda + 1}}{\lambda + 1} = \frac{2^{\lambda + 1} - 1}{\lambda + 1}. \]

### Step 2: Substitute the Integral into the Limit
Now, we substitute \(I(\lambda)\) back into the limit:

\[ \lim_{\lambda \to 0} \left( \frac{2^{\lambda + 1} - 1}{\lambda + 1} \right)^{\frac{1}{\lambda}}. \]

### Step 3: Analyze the Limit
As \(\lambda \to 0\), both the numerator \(2^{\lambda + 1} - 1\) and the denominator \(\lambda + 1\) approach 1. Thus, we have an indeterminate form of the type \(1^{\infty}\).

### Step 4: Apply the Exponential Limit Rule
To resolve the indeterminate form, we can use the property:

\[ \lim_{\lambda \to 0} a^{g(\lambda)} = e^{\lim_{\lambda \to 0} (f(\lambda) - 1) g(\lambda)}, \]

where \(f(\lambda) = 2^{\lambda + 1} - 1\) and \(g(\lambda) = \frac{1}{\lambda + 1}\).

### Step 5: Calculate the Limit Inside the Exponential
Now we need to compute:

\[ \lim_{\lambda \to 0} \left( 2^{\lambda + 1} - 1 \right) \cdot \frac{1}{\lambda + 1}. \]

Using the fact that \(2^{\lambda + 1} = 2 \cdot 2^{\lambda}\), we can expand \(2^{\lambda}\) using the Taylor series:

\[ 2^{\lambda} \approx 1 + \lambda \ln(2) \quad \text{as } \lambda \to 0. \]

Thus,

\[ 2^{\lambda + 1} - 1 \approx 2(1 + \lambda \ln(2)) - 1 = 2 + 2\lambda \ln(2) - 1 = 1 + 2\lambda \ln(2). \]

Now substituting this back, we have:

\[ \lim_{\lambda \to 0} \frac{1 + 2\lambda \ln(2) - 1}{\lambda + 1} = \lim_{\lambda \to 0} \frac{2\lambda \ln(2)}{\lambda + 1} = 2 \ln(2) \cdot \lim_{\lambda \to 0} \frac{\lambda}{\lambda + 1} = 2 \ln(2) \cdot 0 = 0. \]

### Step 6: Final Calculation
Thus, we have:

\[ \lim_{\lambda \to 0} \left( 2^{\lambda + 1} - 1 \right) \cdot \frac{1}{\lambda + 1} = 0. \]

Therefore,

\[ \lim_{\lambda \to 0} \left( \int_{0}^{1} (1+x)^{\lambda} \, dx \right)^{\frac{1}{\lambda}} = e^{0} = 1. \]

### Final Result
The final result is:

\[ \lim_{\lambda \to 0} \left( \int_{0}^{1} (1+x)^{\lambda} \, dx \right)^{\frac{1}{\lambda}} = 1. \]