Question

Find the limit:

lim as t → infinity (sqr t + t^2) / ( 5t − t^2)

Answer 1

Divide everything by \(t^2\) since if is the highest degree term in the denominator.

Answer 2

use l'hopitals rule

Answer 3

is it
\[\frac{\sqrt{t+t^2}}{5t-t^2}\] or is it
\[\frac{\sqrt{t}+t^2}{5t-t^2}\]

Answer 4

the second one

Answer 5

then use nothing
it is the ratio of the leading coefficients

Answer 6

the first time you do the limit it goes to the form inf/inf so use l'hopitals

Answer 7

I get ( sqr t ) / (5t) is that the answer?

Answer 8

oh heck no
you don't use any calculus for this

Answer 9

No calculus for limits?

Answer 10

and also @latona heck no
it is the ratio of the leading coefficients

Answer 11

this is what i got when I divided by t^2

Answer 12

no, not for limits as \(x\) goes to infinity, which is a synonym for horizontal asymptotes

Answer 13

the leading coefficient of the numerator is 1, the leading coefficient of the denominator is -1, and the ration is \(\frac{1}{-1}=-1\)

Answer 14

*ratio

Answer 15

Kind of confused now how to approach the problem. Can you help further, anyone? Thanks.

Answer 16

do you know what l'hopitals rule is?

Answer 17

no

Answer 18

Do you know how to divide?

Answer 19

Divide everything by \(t^2\) like I said.

Answer 20

when I divide everything by t^2, I get \[\frac{ \sqrt{t} }{ 5t }\]

Answer 21

the numerator has \(t^2\) and the denominator as \(-t^2\)
the other stuff is unimportant since the degrees are all smaller than 2
\[\frac{t^2}{-t^2}=-1\] is your asymptote