Calculus

I am maybe overthinking this, but what is the lim as n-> infinity of |(n+1)/(n+2)|

?

I am trying to use the ratio test to find interval of convergence for the infiinite series (n/n+1)((-2x)^(n-1))
so i did the limit of |(n+1)/(n+2)| but i am wondering if the limit is simply =1, or if it is infinity/infinity so use l'hopital's rule???
maybe i am overthinking it and the limit is just 1.

1. the limit of your problem is 1
as n --> infinity , both numerators and denominator are practiacally the same, with the denominator being one less than the numerator.
e.g. if n = 1 000 000
then we have 1 000 001/1 000 002 , pretty close to 1 wouldn't you say?

posted by Reiny
2. since (n+1)/(n+2) = 1 - 1/(n+2) it is clear that as n grows, the value approaches 1.

Yes, L'Hopital's Rule also works, since you have infinity/infinity. taking derivatives gives 1/1 as the limit.

posted by Steve

Similar Questions

1. Calc. Limits

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim
2. calc

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim

Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim
4. Calculus

Find the horizontal asymptote of f(x)=e^x - x lim x->infinity (e^x)-x= infinity when it's going towards infinity, shouldn't it equal to negative infinity, since 0-infinity = - infinity lim x-> -infinity (e^x)-x= infinity
5. calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me
6. calculus - interval of convergence

infinity of the summation n=0: ((n+2)/(10^n))*((x-5)^n) .. my work so far. i used the ratio test = lim (n-->infinity) | [((n+3)/(10^(n+1)))*((x-5)^(n+1))] / [((n+2)/(10^n))*((x-5)^n)] | .. now my question is: was it ok for me
7. Pre-cal

Please determine the following limits if they exist. If the limit does not exist put DNE. lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity lim 4n-3 / 3n^2+2 n-> infinity I did lim 2+6x-3x^2 / (2x+1)^2 x-> - infinity
8. Math

1. If -1/infinity = infinity or -infinity ? 2. If lim x->infinity^- = infinity & lim x->inifinity^+ = -infinity, then lim x->infinity = does not exist. Am i right? If im wrong please tell me the reason why?
9. Math

1. If -1/infinity = infinity or -infinity ? 2. If lim x->infinity^- = infinity & lim x->inifinity^+ = -infinity, then lim x->infinity = does not exist. Am i right? If im wrong please tell me the reason why?
10. calculus

State which of the conditions are applicable to the graph of y = f(x). (Select all that apply.) lim x→infinity f(x) = −infinity lim x→a+ f(x) = L lim x→infinity f(x) = L f is continuous on [0, a] lim

More Similar Questions