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

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(C) We need to evaluate $\mathop {\lim }\limits_{n 	o \infty } \cos \left( {\pi \sqrt {{n^2} + n} } ight)$.Since $\cos(x) = \cos(x - n\pi + n\pi) =

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(C) We need to evaluate $\mathop {\lim }\limits_{n 	o \infty } \cos \left( {\pi \sqrt {{n^2} + n} } ight)$.Since $\cos(x) = \cos(x - n\pi + n\pi) = \cos(n\pi) \cos(x - n\pi) + \sin(n\pi) \sin(x - n\pi) = (-1)^n \cos(x - n\pi)$,we consider:$\mathop {\lim }\limits_{n 	o \infty } (-1)^n \cos \left( {\pi \sqrt {{n^2} + n} - n\pi } ight)$$= \mathop {\lim }\limits_{n 	o \infty } (-1)^n \cos \left( {\pi \left( \sqrt {{n^2} + n} - n ight)} ight)$Rationalizing the expression inside the cosine:$= \mathop {\lim }\limits_{n 	o \infty } (-1)^n \cos \left( {\pi \frac{{n^2 + n - n^2}}{{\sqrt {{n^2} + n} + n}}} ight)$$= \mathop {\lim }\limits_{n 	o \infty } (-1)^n \cos \left( {\pi \frac{n}{{n(\sqrt {1 + 1/n} + 1)}}} ight)$$= \mathop {\lim }\limits_{n 	o \infty } (-1)^n \cos \left( {\frac{\pi}{{\sqrt {1 + 1/n} + 1}}} ight)$As $n 	o \infty$,the argument of cosine approaches $\frac{\pi}{2}$.Thus,the limit is $\mathop {\lim }\limits_{n 	o \infty } (-1)^n \cos \left( \frac{\pi}{2} ight) = \mathop {\lim }\limits_{n 	o \infty } (-1)^n 	imes 0 = 0$.

$\mathop {\lim }\limits_{n \to \infty } \cos \left( {\pi \sqrt {{n^2} + n} } \right)$,where $n \in I$,is

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