`lim_(n to oo) n/(2^n)int_0^2x^n \ dx` equals

Explanation

To solve the limit \( L = \lim_{n \to \infty} \frac{n}{2^n} \int_0^2 x^n \, dx \), we will follow these steps:

### Step 1: Evaluate the integral
The integral \( \int_0^2 x^n \, dx \) can be computed using the formula for the integral of \( x^n \):
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
Thus,
\[ \int_0^2 x^n \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^2 = \frac{2^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{2^{n+1}}{n+1} \]

### Step 2: Substitute the integral back into the limit
Now we substitute the result of the integral back into the limit:
\[ L = \lim_{n \to \infty} \frac{n}{2^n} \cdot \frac{2^{n+1}}{n+1} \]
This simplifies to:
\[ L = \lim_{n \to \infty} \frac{n \cdot 2^{n+1}}{2^n (n+1)} = \lim_{n \to \infty} \frac{n \cdot 2}{n+1} \]

### Step 3: Simplify the expression
We can simplify the expression further:
\[ L = 2 \cdot \lim_{n \to \infty} \frac{n}{n+1} \]

### Step 4: Evaluate the limit
Now, we evaluate the limit:
\[ \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1 \]
Thus,
\[ L = 2 \cdot 1 = 2 \]

### Final Answer
Therefore, the value of the limit is:
\[ \boxed{2} \]