If `0 lt alpha lt beta` then `lim_(n to oo) (beta^(n) + alpha^(n))^((1)/(n))` is equal to

Options

A. `alpha`
B. `beta`
C. `alpha beta`
D. e

Answer

To solve the limit \( \lim_{n \to \infty} ( \beta^n + \alpha^n )^{\frac{1}{n}} \) where \( 0 < \alpha < \beta \), we can follow these steps:

### Step 1: Factor out \( \beta^n \)
We start by factoring \( \beta^n \) out of the expression inside the limit:

\[ \lim_{n \to \infty} ( \beta^n + \alpha^n )^{\frac{1}{n}} = \lim_{n \to \infty} \left( \beta^n \left( 1 + \left( \frac{\alpha}{\beta} \right)^n \right) \right)^{\frac{1}{n}} \]

### Step 2: Simplify the expression
Using the property of exponents, we can simplify the expression:

\[ = \lim_{n \to \infty} \left( \beta^n \right)^{\frac{1}{n}} \left( 1 + \left( \frac{\alpha}{\beta} \right)^n \right)^{\frac{1}{n}} \]

This simplifies to:

\[ = \lim_{n \to \infty} \beta \left( 1 + \left( \frac{\alpha}{\beta} \right)^n \right)^{\frac{1}{n}} \]

### Step 3: Analyze the term \( \left( \frac{\alpha}{\beta} \right)^n \)
Since \( 0 < \alpha < \beta \), we have \( \frac{\alpha}{\beta} < 1 \). Therefore, as \( n \) approaches infinity, \( \left( \frac{\alpha}{\beta} \right)^n \) approaches 0:

\[ \lim_{n \to \infty} \left( \frac{\alpha}{\beta} \right)^n = 0 \]

### Step 4: Substitute back into the limit
Now substituting this back into our limit gives us:

\[ = \lim_{n \to \infty} \beta \left( 1 + 0 \right)^{\frac{1}{n}} \]

### Step 5: Evaluate the limit
As \( n \) approaches infinity, \( \left( 1 + 0 \right)^{\frac{1}{n}} \) approaches 1:

\[ = \beta \cdot 1 = \beta \]

### Final Result
Thus, the limit is:

\[ \lim_{n \to \infty} ( \beta^n + \alpha^n )^{\frac{1}{n}} = \beta \]