mathop {Lim}limits_{n to infty } cos left( {p | vedclass

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(C) We want to evaluate $L = \mathop {Lim}\limits_{n 	o \infty } \cos \left( {\pi \sqrt {{n^2} + n} } ight)$.Since $n$ is an integer,we can write $

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is equal to $0$

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(C) We want to evaluate $L = \mathop {Lim}\limits_{n 	o \infty } \cos \left( {\pi \sqrt {{n^2} + n} } ight)$.Since $n$ is an integer,we can write $\cos \left( {\pi \sqrt {{n^2} + n} } ight) = \cos \left( {n\pi - n\pi + \pi \sqrt {{n^2} + n} } ight) = \cos \left( {n\pi - \pi \left( {n - \sqrt {{n^2} + n} } ight)} ight)$.Using the identity $\cos(n\pi - 	heta) = (-1)^n \cos(	heta)$,we have $L = \mathop {Lim}\limits_{n 	o \infty } (-1)^n \cos \left( {\pi \left( {n - \sqrt {{n^2} + n} } ight)} ight)$.Now,rationalize the expression inside the cosine:$n - \sqrt {{n^2} + n} = \frac{{(n - \sqrt {{n^2} + n} )(n + \sqrt {{n^2} + n} )}}{{n + \sqrt {{n^2} + n} }} = \frac{{{n^2} - ({n^2} + n)}}{{n + \sqrt {{n^2} + n} }} = \frac{{ - n}}{{n + n\sqrt {1 + 1/n} }} = \frac{{ - 1}}{{1 + \sqrt {1 + 1/n} }}$.As $n 	o \infty$,this expression approaches $\frac{{ - 1}}{{1 + 1}} = -1/2$.Thus,$L = \mathop {Lim}\limits_{n 	o \infty } (-1)^n \cos \left( {\pi \left( -\frac{1}{2} ight)} ight) = \mathop {Lim}\limits_{n 	o \infty } (-1)^n \cos \left( -\frac{\pi}{2} ight) = \mathop {Lim}\limits_{n 	o \infty } (-1)^n \cdot 0 = 0$.

$\mathop {Lim}\limits_{n \to \infty } \cos \left( {\pi \sqrt {{n^2} + n} } \right)$ when $n$ is an integer:

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