11it is a debatable ques
but the concept is clear that cos x is a function whose a value lies between -1and 1
and ∞ is not a number so cos ∩√n√1+n2 where lim n->∞
the limit does not exist (D)
15 Answers
11
11Reason
Using L-Hospitals rule even if infinitely differentiated still sin or cos will be present. Therefore It is D)
62virang.. i did not unerstand how you used LH rule here
I think it will be by this logic
cos(\pi n \sqrt{1+1/n}) \\ \text{approaches }cos(\pi (n +1/2)) \\ =0
11i feel -1
q=cos(ΠLtn→∞√n2+n)
applying LH
=cos(Π(1/2)2n+1/√n2+n)
so -1
62basically what i am saying is that
√ n2 + n approaches n+1/2 as n approaches infinity
prove this.. you will be done :)
11basically what i am saying is that
√ n2 + n approaches n+1/2 as n approaches infinity
prove this.. you will be done :)
SIR NAHIN SAMAJH AA RAHA YEH KAHAN SE AAYA[2][17]
1@Mr.Manipal,
\sqrt{n^{2}+n}=\sqrt{(n+\frac{1}{2})^{2}-(\frac{1}{2})^{2}}
62yes i think in the question part where he aki removed the right part, n is an integer is given..
otherwise it will be a wasted good question ;)