If `f(x)=x+1`. Find `lim_(nrarroo)1/n(1+f(5/n)+f(10/n)+. . . + f(5((n-1)/(n)))`

Answer

To solve the limit problem, we start with the function \( f(x) = x + 1 \) and need to evaluate:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + f\left(\frac{5}{n}\right) + f\left(\frac{10}{n}\right) + \ldots + f\left(\frac{5(n-1)}{n}\right) \right) \]

### Step 1: Rewrite the expression using the function \( f(x) \)

We can express the terms inside the limit using the definition of \( f(x) \):

\[ f\left(\frac{5k}{n}\right) = \frac{5k}{n} + 1 \]

for \( k = 0, 1, 2, \ldots, n-1 \).

### Step 2: Substitute into the limit

Substituting this into our limit gives:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \left(\frac{5 \cdot 0}{n} + 1\right) + \left(\frac{5 \cdot 1}{n} + 1\right) + \ldots + \left(\frac{5(n-1)}{n} + 1\right) \right) \]

This simplifies to:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \sum_{k=0}^{n-1} \left(\frac{5k}{n} + 1\right) \right) \]

### Step 3: Simplify the summation

The summation can be split into two parts:

\[ \sum_{k=0}^{n-1} \left(\frac{5k}{n} + 1\right) = \sum_{k=0}^{n-1} \frac{5k}{n} + \sum_{k=0}^{n-1} 1 \]

The second summation is simply \( n \):

\[ \sum_{k=0}^{n-1} 1 = n \]

The first summation can be calculated as follows:

\[ \sum_{k=0}^{n-1} \frac{5k}{n} = \frac{5}{n} \sum_{k=0}^{n-1} k = \frac{5}{n} \cdot \frac{(n-1)n}{2} = \frac{5(n-1)}{2} \]

### Step 4: Combine the results

Now, substituting back, we have:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \frac{5(n-1)}{2} + n \right) \]

This simplifies to:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \frac{5(n-1)}{2} + n \right) = \lim_{n \to \infty} \frac{1}{n} \left( n + \frac{5(n-1)}{2} + 1 \right) \]

### Step 5: Factor out \( n \)

Factoring out \( n \) gives:

\[ \lim_{n \to \infty} \left( 1 + \frac{5(n-1)}{2n} + \frac{1}{n} \right) = \lim_{n \to \infty} \left( 1 + \frac{5}{2} - \frac{5}{2n} + \frac{1}{n} \right) \]

As \( n \to \infty \), the terms \( -\frac{5}{2n} \) and \( \frac{1}{n} \) approach 0. Thus, we have:

\[ 1 + \frac{5}{2} = \frac{7}{2} \]

### Final Result

Therefore, the limit evaluates to:

\[ \boxed{\frac{7}{2}} \]