Question

Infinite limits

Answer 1

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

Answer 2

ok im wonderring how i show my work, my first thought would be to mulitply by the conjugate

Answer 3

not zero in any case

Answer 4

and yes, multiplying top and bottom by the conjugate would be a good idea

Answer 5

should be pretty clear that there will be an \(a\) and \(b\) in your answer

Answer 6

multiplying by the conjugate should work well

you'll end up with the numerator going faster than the denominator when x -> inf, so the answer would be infinity

Answer 7

really? so suppose since \(\lim_{x\to \infty}x=\infty\) and \(\lim_{x\to \infty}x^2=\infty\) that \(\lim_{x\to \infty}(x-x^2)=0\)

Answer 8

wow lets right it out

Answer 9

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

Answer 10

got it

Answer 11

thats after conjugate

Answer 12

no i would like to mulitply num/dom by x^-2

Answer 13

but i dont think i can do that

Answer 14

now we can do it with our eyeballs
denonminator grows like \(2x\) and numerator like \((a-b)x\) so your answer is \[\frac{a-b}{2}\] not zero (unless \(a=b\)) , and not infinity

Answer 15

ok fine

Answer 16

the tedious math teacher way is to divide top and bottom by\(x\) which means of course in radical it will look like \(x^2\) and get
\[\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}\]

Answer 17

then take the limit as \(x\to \infty\) and get \(\frac{a-b}{2}\)
steps clear?

Answer 18

@satellite73 that leaves me with a-b/2x

Answer 19

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

Answer 20

\[\frac{ a/x-b/x }{ \sqrt{1}+\sqrt{1} }\]

Answer 21

\[\frac{ a-b }{ 2x}\]

Answer 22

@satellite73 you there?

Answer 23

@hartnn can you help me with something real quick/?

Answer 24

hey hartnn

Answer 25

ive got this all worked out, its just at the end i have a leftover x

Answer 26

when u have @satellite73 u don't need anyone!!

Answer 27

i tried calling sat back but no answer :(

Answer 28

u don't have x in the denominator of numerator

Answer 29

u are dividing by x, not by x^2

Answer 30

(a-b) / (root1 + root 1)

Answer 31

ya, but how do i clear the x^2 from the roots in the denom?

Answer 32

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

Answer 33

\(\large \frac{ ax-bx }{ \sqrt{x^2+ax}+\sqrt{x^2+bx} }\frac{ x^{-1} }{ (x^{-1}) }=\frac{a-b}{\sqrt{1+a/x}+\sqrt{1+b/x}}\)

Answer 34

so what is acutally happening when you multiply by x^-1

Answer 35

x^2/x = x

Answer 36

\(\huge \sqrt{x^2+ax}=\sqrt{x^2(1+a/x)}=x\sqrt{(1+a/x)}\)

Answer 37

went outside, but u get how x comes outside ?

Answer 38

are you factoring out sqrt(x)?

Answer 39

first factoring x^2, then taking it out from sqrt, to get only x

Answer 40

can i do this?

Answer 41

\[\sqrt{x^2+ax}=\sqrt{x}\sqrt{x+a}\]

Answer 42

yes! u can.
but here thats of no use.

Answer 43

i dont know how you are doing the other way, but if i do this then multiplying by x^-1 on both top/bottom then i would get a-b/2

Answer 44

u just factored, root x and u multiplying , x^-1
so x^-1 * x^1/2 won't cancel out....

Answer 45

\[\frac{ ax-bx }{ \sqrt{x}\sqrt{x+a}+\sqrt{x}\sqrt{x+b} }(\frac{ x^{-1} }{ x^{-1} })\]

Answer 46

\(\sqrt x x^{-1}= ? \)

Answer 47

1

Answer 48

NO!!

Answer 49

damn im losing it

Answer 50

x x^-1 = 1

Answer 51

how do i clear a square root like that?

Answer 52

\(\large \sqrt{x^2+ax}=\sqrt{x^2(1+a/x)}=x\sqrt{(1+a/x)}\)

Answer 53

ok, i think i got it now

Answer 54

if a doesnt have another x to give it becomes a/x

Answer 55

then u multiply by x^-1
\(\\huge =x^{-1}x\sqrt{(1+a/x)}=\sqrt{(1+a/x)}\)

Answer 56

yes. thats correct

Answer 57

ok thanks!!!!!!!!!!!!

Answer 58

;D

Answer 59

welcome :)