Question

The sequence is given by the following equation: t = (5 - n^2)/3n , where n E N [an element of a natural number]. Does this sequence have a limit of n approaches ∞?

Answer 1

You have:
\[\lim_{n \rightarrow \infty} \frac{(5-n^2)}{3n}\].
Divide the whole thing through by n^2 (or do l'hospital)
\[\lim_{n \rightarrow \infty} \frac{\frac{5}{n^2}-1}{\frac{3}{n}}\].
Then you have:
\[(1/3)\lim_{n \rightarrow \infty}\frac{5}{n}-n=(1/3)(0-\infty)=-\infty \rightarrow D.N.E.\].

Answer 2

lim (5/3n) =0 as n tends to infinity (n gets large 5/3n gets very very small)
lim (-n^2/3n)=-n/3 which has no limit as n can be as large as we like.
lim (5/3n) +lim -n^2/3n = 0 -(n/3)
adding the limits no the sum is divergent and has no limit.

Answer 3

this equation doesn't hav a limit. this is simply explained by the fact that the power of n in the numerator is higher compared to the one in denominator....hence for a sequence to converge and hav a limit, the denominator will always be>=numerator.
understood?

Answer 4

with your eyeball. degree of numerator is bigger than degree of denominator, limit is
\[-\infty\] otherwise known as no limit