Question

ljkh;lkj

Answer 1

@jim_thompson5910

Answer 2

as x->inf, what's happening to x^2+ax?

Answer 3

its getting super big

Answer 4

yes, so the square root of something big (that keeps growing) is also something that keeps growing

as x->inf, then sqrt(x^2+ax) -> inf

Answer 5

same with sqrt(x^2+bx)

Answer 6

so we have something of the form infinity - infinity, which is an indeterminate form

Answer 7

what do we do with indeterminate forms?

Answer 8

thats the part im blanking on

Answer 9

one sec

Answer 10

ok

Answer 11

ok I found this page

http://www.vitutor.com/calculus/limits/minus_infinity.html

Answer 12

and it's recommending you multiply and divide the expression by the conjugate

Answer 13

ok. what would the conjugate be?

Answer 14

never mind i got it

Answer 15

i think

Answer 16

it would be \[\Large \sqrt{x^2+ax}+\sqrt{x^2+bx}\]

Answer 17

it looks like
\[\infty-\infty\] at the moment
multiply by what @jim_thompson5910 said and you will get it

Answer 18

((sqrt(x^2+ax)-sqrt(x^2+bx))(sqrt(x^2+ax)+sqrt(x^2+bx))=ax-bx. how would i continue?

Answer 19

you should have this after you multiply by

\[\Large \frac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\]

see attached pdf

Answer 20

yeah thats what i have

Answer 21

once you get to

\[\large \frac{ax-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\]

you can evaluate the limit as x --> infinity. You'll then get the form infinity/infinity

That's also indeterminate, but you can at least use L'Hospital's Rule to find the actual limit (if it exists)

Answer 22

i havent learned L'hospitals rule and it should exist

Answer 23

ok one sec

Answer 24

you can multiply top and bottom by 1/x

Answer 25

you can or you can just look with your eyeballs

Answer 26

since you heading towards +infinity that won't pose a problem

Answer 27

\[\lim_{x\to \infty}\large \frac{ax-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\]
\[=\frac{(a-b)x}{2\sqrt{x^2}}=\frac{(a-b)x}{2x}=\frac{a-b}{2}\]

Answer 28

thank you @jim_thompson5910 and @satellite73 and @xapproachesinfinity

Answer 29

xapproachesinfinity and satellite73 have the right idea and it's shown in this pdf

Answer 30

my way would be doing \(x=\sqrt{x^2}\) in the bottom since x is positive
if it is not then be careful
the result is what @satellite73 got

Answer 31

the PDF showed everything^_^