1. Limit n--> infinity ( 2 - (1/n) + (3/n^2) i got this to be 2
2. Limit n--> infinity (1+2n)/ (1-3n) i got this to be one
3. liminit n--> infnity of (1.1)^ n im not sure what this is, please help
4. Limitnit n--> infinity (3^n/5^n) = 3/5 ?
pplease help me thank you soo much :D
2. Limit n--> infinity (1+2n)/ (1-3n) i got this to be one
3. liminit n--> infnity of (1.1)^ n im not sure what this is, please help
4. Limitnit n--> infinity (3^n/5^n) = 3/5 ?
pplease help me thank you soo much :D
-
1. Limit n--> infinity ( 2 - (1/n) + (3/n^2) i got this to be 2
Solution
lim (2 - 1/n + 3/n²) does indeed equal 2 since other terms go to zero
n→∞.......as n goes to infinity
2. Limit n--> infinity (1+2n)/ (1-3n) i got this to be one
.......1+2n
lim ———— IS NOT ONE
n→∞..1-3n
Note that n is going to infinity here and this limit
becomes equivalent to
.......1/n+2
lim ———— and that convergest to -2/3 as n→∞
n→∞..1/n-3
3. liminit n--> infnity of (1.1)^ n im not sure what this is please help
...........n
lim (1.1).... is going to be infinity so the limit does not exist
n→∞..
4. Limitnit n--> infinity (3^n/5^n) = 3/5 ?
.........3^n
lim ————
n→∞..5^n
equals
............n
lim (3/5) and because 3/5<1 as n→∞ this limit converges to ZERO
n→∞.
Solution
lim (2 - 1/n + 3/n²) does indeed equal 2 since other terms go to zero
n→∞.......as n goes to infinity
2. Limit n--> infinity (1+2n)/ (1-3n) i got this to be one
.......1+2n
lim ———— IS NOT ONE
n→∞..1-3n
Note that n is going to infinity here and this limit
becomes equivalent to
.......1/n+2
lim ———— and that convergest to -2/3 as n→∞
n→∞..1/n-3
3. liminit n--> infnity of (1.1)^ n im not sure what this is please help
...........n
lim (1.1).... is going to be infinity so the limit does not exist
n→∞..
4. Limitnit n--> infinity (3^n/5^n) = 3/5 ?
.........3^n
lim ————
n→∞..5^n
equals
............n
lim (3/5) and because 3/5<1 as n→∞ this limit converges to ZERO
n→∞.
-
1) correct
2) wrong, you may apply H'lopitals rule (take the derivative of the top, divided by the derivative of the bottom and you get -2/3 to be correct
3) as n goes to infinity, it will also be infinity, since the number out in front is great than 1, this one would be does not exist
4) wrong, this can be re-written as (3/5)^n, since the number in front of n is between -1 and 1, we can say that the limit goes to 0 (the reason you can say if the number in front is between -1 and 1 it goes to 0 is because the denominator is larger than the numerator, meaning if you took 3^10 / 5^10, the denominator is much bigger, making it approach 0 as n goes to infinity)
Hope this helps :)
2) wrong, you may apply H'lopitals rule (take the derivative of the top, divided by the derivative of the bottom and you get -2/3 to be correct
3) as n goes to infinity, it will also be infinity, since the number out in front is great than 1, this one would be does not exist
4) wrong, this can be re-written as (3/5)^n, since the number in front of n is between -1 and 1, we can say that the limit goes to 0 (the reason you can say if the number in front is between -1 and 1 it goes to 0 is because the denominator is larger than the numerator, meaning if you took 3^10 / 5^10, the denominator is much bigger, making it approach 0 as n goes to infinity)
Hope this helps :)
-
Yes and also was anyone else reminded of mean girls by this question??
-
yes its right