The value of `lim_(nrarroo)((1)/(2n)+(1)/(2n+1)+(1)/(2n+2)+…..+(1)/(4n))` is equal to

Options

A. `e^(2)`
B. ln 2
C. ln 4
D. 3 ln 2

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} + \ldots + \frac{1}{4n} \right), \]

we will follow these steps:

### Step 1: Identify the terms in the summation
The summation consists of terms from \(\frac{1}{2n}\) to \(\frac{1}{4n}\). The number of terms can be calculated as follows:

- The first term is \(\frac{1}{2n}\) (when \(r=0\)).
- The last term is \(\frac{1}{4n}\) (when \(r=2n\)).

Thus, the general term can be expressed as:

\[ \frac{1}{2n + r} \quad \text{for } r = 0, 1, 2, \ldots, 2n. \]

### Step 2: Rewrite the limit as a summation
We can rewrite the limit as:

\[ \lim_{n \to \infty} \sum_{r=0}^{2n} \frac{1}{2n + r}. \]

### Step 3: Factor out \(\frac{1}{n}\)
To facilitate the evaluation, we can factor out \(\frac{1}{n}\) from the denominator:

\[ \sum_{r=0}^{2n} \frac{1}{2n + r} = \sum_{r=0}^{2n} \frac{1/n}{2/n + r/n} = \frac{1}{n} \sum_{r=0}^{2n} \frac{1}{\frac{2}{n} + \frac{r}{n}}. \]

### Step 4: Convert the summation to an integral
As \(n\) approaches infinity, the sum can be approximated by an integral. Let \(t = \frac{r}{n}\), then \(dt = \frac{1}{n}\) and the limits change from \(0\) to \(2\):

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n} \frac{1}{\frac{2}{n} + \frac{r}{n}} \approx \int_{0}^{2} \frac{1}{2 + t} dt. \]

### Step 5: Evaluate the integral
Now we compute the integral:

\[ \int_{0}^{2} \frac{1}{2 + t} dt. \]

The antiderivative of \(\frac{1}{2 + t}\) is \(\ln |2 + t|\). Thus, we evaluate:

\[ \left[ \ln(2 + t) \right]_{0}^{2} = \ln(4) - \ln(2) = \ln\left(\frac{4}{2}\right) = \ln(2). \]

### Final Result
Therefore, the value of the limit is:

\[ \lim_{n \to \infty} \left( \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} + \ldots + \frac{1}{4n} \right) = \ln(2). \]