Evaluate : `lim_(n to oo)[(sqrt(n))/((3+4sqrt(n))^(2))+(sqrt(n))/(sqrt(2)(3sqrt(2)+4sqrt(n))^(2))+(sqrt(n))/(sqrt(3)(3sqrt(3)+4sqrt(n))^(2))+.......+(1)/(49n)]`

Options

A. `(1)/(14)`
B. `(1)/(16)`
C. `(1)/(18)`
D. `(1)/(20)`

Answer

To evaluate the limit

\[ \lim_{n \to \infty} \left[ \frac{\sqrt{n}}{(3 + 4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2} + 4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{3}(3\sqrt{3} + 4\sqrt{n})^2} + \ldots + \frac{1}{49n} \right], \]

we can break down the problem step by step.

### Step 1: Rewrite the limit expression
The expression can be rewritten in summation form:

\[ \lim_{n \to \infty} \sum_{r=1}^{49} \frac{\sqrt{n}}{\sqrt{r} (3\sqrt{r} + 4\sqrt{n})^2}. \]

### Step 2: Factor out \(\sqrt{n}\)
We can factor \(\sqrt{n}\) out of the denominator:

\[ = \lim_{n \to \infty} \sum_{r=1}^{49} \frac{\sqrt{n}}{\sqrt{r} \cdot 4^2 n} \cdot \frac{1}{\left(\frac{3\sqrt{r}}{4\sqrt{n}} + 1\right)^2}. \]

This simplifies to:

\[ = \lim_{n \to \infty} \sum_{r=1}^{49} \frac{1}{\sqrt{r} \cdot 16} \cdot \frac{1}{\left(\frac{3\sqrt{r}}{4\sqrt{n}} + 1\right)^2}. \]

### Step 3: Analyze the limit as \(n \to \infty\)
As \(n\) approaches infinity, \(\frac{3\sqrt{r}}{4\sqrt{n}} \to 0\). Thus, we have:

\[ \left(\frac{3\sqrt{r}}{4\sqrt{n}} + 1\right)^2 \to 1^2 = 1. \]

### Step 4: Evaluate the limit
Now we can evaluate the limit:

\[ = \sum_{r=1}^{49} \frac{1}{\sqrt{r} \cdot 16} \cdot 1 = \frac{1}{16} \sum_{r=1}^{49} \frac{1}{\sqrt{r}}. \]

### Step 5: Calculate the sum
The sum \(\sum_{r=1}^{49} \frac{1}{\sqrt{r}}\) can be approximated or computed directly. For large \(r\), this behaves like an integral, but since we are summing up to 49, we can compute it directly or use numerical methods to find:

\[ \sum_{r=1}^{49} \frac{1}{\sqrt{r}} \approx 8.5 \quad \text{(using numerical approximation)}. \]

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

\[ \lim_{n \to \infty} \left[ \frac{\sqrt{n}}{(3 + 4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2} + 4\sqrt{n})^2} + \ldots + \frac{1}{49n} \right] \approx \frac{1}{16} \cdot 8.5 = \frac{8.5}{16} = \frac{17}{32}. \]

### Conclusion
The final limit evaluates to:

\[ \frac{17}{32}. \]