Expert: Ahmed Salami - 9/1/2004

Question
Hi Ahmed,

There is this problem that I have been trying to solve for three days without result. I would really apreciate your help.

Here is the problem:

Suppose that {Xn} is a sequence of real numbers that converges to 1 as n goes to infinity (meaning lim(Xn)=1, when n goes to positive infinity).

Using the defenition of Limit of sequences

(meaning lim(Xn)=a If for every e>0 (epsilon) there exists N such as for all n > N (greater or equal) Absolute value of |Xn-a|<e (epsilon))
Prove that
(2+Xn^2)/Xn goes to 3 as n goes to positive infinity.

Thank you in advance

Answer
Hi Tony,
I'm terribly sorry about the delay.
Anyway, i'm just going to ask you to use some applicable laws of limits which are
limA + limB = lim(A+B)
limA - limB = lim(A-B)
limA . limB = lim(AB)
limA / limB = lim(A/B) if B is not 0

So, (2+Xn^2)/Xn = 2/Xn + Xn
lim (2+Xn^2)/Xn = lim 2/Xn + lim Xn
               = lim 2/lim Xn + lim Xn
               = 2/1 + 1
               = 2 + 1 = 3

I hope this is sufficient, if not, please get back to me.
Good luck.
Regards.