Question

limit as x approaches infinity

Answer 1

\[\lim_{x \rightarrow \infty}\frac{ x^2+3x+2 }{ x-1 }\]

Answer 2

I know the answer is infinity but I have to do it by dividing every term by x

Answer 3

\[\lim_{x \rightarrow \infty}\frac{ \frac{x^2}{x}+\frac{3x}{x}+\frac{2}{x} }{ \frac{x}{x}-\frac{1}{x} }\]

Answer 4

start by simplifying. x^2/x =x x/x = 1 etc

Answer 5

\[\lim_{x \rightarrow \infty}\frac{ x+3+\frac{2}{x} }{ 1-\frac{1}{x} }\]

Answer 6

\[\lim_{x \rightarrow \infty}\frac{ x+3+0 }{ 1-0}\]

Answer 7

ok so it is just infinity from this point, correct?

Answer 8

yes

Answer 9

same problem but x approaches negative infinity so I would be dividing by negative -x, or not?

Answer 10

im not sure tbh, i don't remember doing this type of thing to evaluate limits--i probably did it but have forgotten in the 5 years since my last calc class lol

Answer 11

i mean, probably not and/or it probably doesnt matter. since it goes to \(-\infty:x\rightarrow -\infty\)

Answer 12

ok I think that it is true, thanks :)

Answer 13

I mean really, on this type of thing, it is really obvious that the numerator approaches infty/-infty >> the dominator so its gonna be \[\frac{\pm \infty}{n-\infty}\]

Answer 14

you may try changing it to positive limit :
\(t = -x\)
as \(x \to -\infty \), \(t \to \infty\)

Answer 15

\[\lim_{x \rightarrow -\infty}\frac{ x^2+3x+2 }{ x-1 } = \lim_{t \rightarrow \infty}\frac{ t^2-3t+2 }{ -t-1 }\]