mathop {lim }limits_{n to infty } frac{1}{{{n | vedclass

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(D) The given limit is $L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{{{n^2}}}\sum\limits_{r = 1}^{2n} r \cos \left( \frac{{{r^2}}}{{{n^2}}} i

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$\frac{{\sin 4}}{2}$

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(D) The given limit is $L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{{{n^2}}}\sum\limits_{r = 1}^{2n} r \cos \left( \frac{{{r^2}}}{{{n^2}}} ight)$.We can rewrite this as a Riemann sum by setting $x = \frac{r}{n}$,so $dx = \frac{1}{n}$.As $n 	o \infty$,the sum becomes the definite integral $\int_0^2 x \cos(x^2) dx$.Let $u = x^2$,then $du = 2x dx$,which implies $x dx = \frac{1}{2} du$.When $x = 0$,$u = 0$. When $x = 2$,$u = 4$.Substituting these into the integral,we get $L = \int_0^4 \cos(u) \cdot \frac{1}{2} du$.$L = \frac{1}{2} [\sin(u)]_0^4 = \frac{1}{2} (\sin 4 - \sin 0) = \frac{\sin 4}{2}$.

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}}\left[ {1\cos \frac{1}{{{n^2}}} + 2\cos \frac{4}{{{n^2}}} + 3\cos \frac{9}{{{n^2}}} + .... + 2n\cos 4} \right]$ is equal to-

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