Question

How to tell if this limit approaches +/- infinity

Answer 1

\[\frac{ 2x^{4} + 6x^{2}+5 }{ 3 + x^3 } \]

Answer 2

lim as x-> infinity

Answer 3

0

Answer 4

@ksaimouli the answer is infinity though

Answer 5

but u need to take the horizontal asymptot

Answer 6

Why?

Answer 7

because it is x->infinity

Answer 8

I still don't understand how you got zero my teacher said it is inifinity

Answer 9

hmm @baldymcgee6 can u help him

Answer 10

he is very good at this i think he will explain u

Answer 11

\[\lim_{x \rightarrow \infty} \frac{ 2x^{4} + 6x^{2}+5 }{ 3 + x^3 } = \infty\]
As x approaches infinity, the numerator becomes exponentially large, in the denominator, the same happens, but the denominator will still be "small" compared to the infinite numerator

Answer 12

So a really big number divided by a smaller number?

Answer 13

When would it be negative infinity? Only when it is approaching negative infinity?

Answer 14

right, you have a large/small, (relatively)
when x-> -infinity, you will notice that all the x values are squared or ^4 which means it will be positive.
http://screencast.com/t/Cq9ShAlg

Answer 15

sorry, the words got cut off, http://screencast.com/t/yxfC6BlwMe

Answer 16

Ohhhh okay ! Thank you so much!

Answer 17

You're welcome :)

Answer 18

btw @ksaimouli there is no horizontal asymptote, but there is an oblique asymptote.

Answer 19

ok

Answer 20

@3psilon i lied!

Answer 21

it looked like to me that the x^3 in the denominator was just x^2, in this case. As x-> -infinity the limit is - infinity as well.

Answer 22

Whoa it matters that much! Okay ! I got cha now :)

Answer 23

if the x is to the power of an odd number, it could be a negative value.
http://screencast.com/t/DlWUwK8Ak45J

Answer 24

Got it got it :) Thank you :)