Question

taking the lim of n^12/ ( n^13- (n-3)^13) as n goes to infinity

Cannot use derivatives . It is a sequence

Answer 1

I cannot use derivatives or hopital's rule inthis problem

Answer 2

@hartnn this is the question. I am trying the squeeze theorem to see if I can get something out of it. I know the limit is 1/39

Answer 3

First answer to :(n-3)^13=?

Answer 4

\[ lim_{ n\rightarrow \infty }\frac{n^{12}}{n^{13}-(n-3)^{13}}\]

Answer 5

What is it ???

Answer 6

@E.ali what do you mean what is it? It is a sequence

Answer 7

Is it your question ?:)

Answer 8

huh ???

Answer 9

my question is how to prove the limit is 1/39 .

Answer 10

use newton binomial expansion

Answer 11

There's a difference between "showing" the limit is 1/39 and "proving" it's 1/39.

The first would involve algebraically manipulating the expression so that finding the limit is simpler, and the second would involve showing that for some \(\epsilon>0\) there is some \(N\) such that \(n\ge N\) implies \(\left|\dfrac{n^{12}}{n^{13}-(n-3)^{13}}-\dfrac{1}{39}\right|<\epsilon\).

Which one is it?

Answer 12

@SithsAndGiggles i. YOu are right. The question says " evaluate the limit or prove it doesn't exists" So I guess showing it is 1/39 is enough

Answer 13

@watchmath it worked I expanded and simplified the denominator , then factor n^12 and took the limit so the denominator went to 3*(13+0+0.......+0) =39 hence 1/39