`Lt_(n rarr oo) [(1)/(n^(2)) sin ((1+ n^(2))/(n^(2))) + (2)/(n^(2)) sin ((4+ n^(2))/(n^(2))) + (3)/(n^(2)) sin ((9+ n^(2))/(n^(2))) + …(2)/(n) sin (5)]`=

Options

A. `cos 2 sin 3`
B. `sin 2 cos 3`
C. `sin 2 sin 3`
D. `cos 2 cos 3`

Answer

To solve the limit

\[ \lim_{n \to \infty} \left[ \frac{1}{n^2} \sin\left(\frac{1+n^2}{n^2}\right) + \frac{2}{n^2} \sin\left(\frac{4+n^2}{n^2}\right) + \frac{3}{n^2} \sin\left(\frac{9+n^2}{n^2}\right) + \ldots + \frac{n}{n^2} \sin\left(\frac{n^2}{n^2}\right) \right], \]

we can rewrite the expression in summation form:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sin\left(\frac{k^2+n^2}{n^2}\right). \]

This can be simplified to:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sin\left(\frac{k^2}{n^2} + 1\right). \]

Next, we can make the substitution \( t = \frac{k}{n} \), which implies \( k = nt \) and \( \Delta k = n \Delta t \). The sum can be approximated by an integral as \( n \to \infty \):

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n^2} \sin\left(\frac{k^2}{n^2} + 1\right) \approx \int_{0}^{1} t \sin(t^2 + 1) \, dt. \]

Now we need to evaluate the integral:

\[ \int_{0}^{1} t \sin(t^2 + 1) \, dt. \]

To solve this integral, we can use the substitution \( u = t^2 + 1 \). Then, \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \). The limits change as follows: when \( t = 0 \), \( u = 1 \) and when \( t = 1 \), \( u = 2 \).

Thus, the integral becomes:

\[ \int_{1}^{2} \frac{1}{2} \sin(u) \, du. \]

Now we can integrate:

\[ \frac{1}{2} \left[-\cos(u)\right]_{1}^{2} = \frac{1}{2} \left[-\cos(2) + \cos(1)\right]. \]

This simplifies to:

\[ \frac{1}{2} \left(\cos(1) - \cos(2)\right). \]

Finally, we can use the cosine subtraction formula:

\[ \cos(1) - \cos(2) = -2 \sin\left(\frac{1 + 2}{2}\right) \sin\left(\frac{2 - 1}{2}\right) = -2 \sin\left(\frac{3}{2}\right) \sin\left(\frac{1}{2}\right). \]

Thus, the limit evaluates to:

\[ \frac{1}{2} \cdot (-2 \sin\left(\frac{3}{2}\right) \sin\left(\frac{1}{2}\right)) = -\sin\left(\frac{3}{2}\right) \sin\left(\frac{1}{2}\right). \]

So the final answer is:

\[ \lim_{n \to \infty} \left[ \frac{1}{n^2} \sin\left(\frac{1+n^2}{n^2}\right) + \frac{2}{n^2} \sin\left(\frac{4+n^2}{n^2}\right) + \frac{3}{n^2} \sin\left(\frac{9+n^2}{n^2}\right) + \ldots + \frac{n}{n^2} \sin\left(\frac{n^2}{n^2}\right) \right] = -\sin\left(\frac{3}{2}\right) \sin\left(\frac{1}{2}\right). \]