Calculate lt [(n+1)^2 - n^2]/(3n+5). n-->infinity.

Expert Answers
hala718 eNotes educator| Certified Educator

lim [(n+1)^2-n^2]/(3n+5)     n--> inf.

= lim (n^2+2n+1-n^2)/(3n+5) = lim (2n+1)/(3n+5)

==> lim (2n+1)/ lim(3n+5) \

Now divide by the highest power (n)

==> lim n(2+1/n)/ lim n(3+5/n)   when n-->inf.

Reduce similar:

==> lim(2+1/n)/lim(3+5/n)   n-->inf.

==> lim (2+0)/ lim(3+0) = 2/3

giorgiana1976 | Student

To calculate the limit of the rational function, when n tends to +inf.,we'll factorize both, numerator and denominator, by the highest power of n, which in this case is n.

We'll have:

lim [(n+1)^2 - n^2]/(3n+5)

To make evident the highest power of n, we'll open the brackets from numerator, first.

lim (n^2+2n+1-n^2)/(3n+5)

e'll reduce the similar terms:

lim (2n+1)/(3n+5)

It is obvious that the highest power of n is n and we'll factorize:

lim n*(2+1/n)/n*(3+5/n)

After reducing similar terms,we'll get:

lim (2+1/n)/(3+5/n) = 2/3

neela | Student

To find the  lt [(n+1)^2 - n^2]/(3n+5). n-->infinity

Solution:

 lt [(n+1)^2 - n^2]/(3n+5). n-->infinity . Simplifying the numerator, we get:

=lt{ n^2+2n+1 - n^2]/(3n+5)

=lt(2n+1)/(3x+5) as n--> infinity. is of the indeteminate form like  infinity/infinity. So we use the L'Hospital's rule of differentiating both numerator and denominator and then take the limit. So,

lt(2n+1)/(3n+5) = (2n+1)'/(3n+5)' = 2/3