Question

limit x approaches infinity, (rad(x^2-9))/(2x-6)

Answer 1

come back @hartnn

Answer 2

lol!
whats rad ?

Answer 3

sqrt

Answer 4

ya

Answer 5

divide x in numerator and denom.

Answer 6

5x-6

Answer 7

rad(x^2-9)
---------- ----> multiply by this
rad(x^2-9)

Answer 8

-y

Answer 9

then what algebraic

Answer 10

r2+s2-t2

Answer 11

x^2-9 : difference of squares

what are the factors?

Answer 12

@Algebraic! is there any need to do that ??
x-> infinity and not 3.

Answer 13

ya whoops.

Answer 14

easy mode

Answer 15

so im here now (x-3)(x+3)/(2x-6)(sqrt(x^2-9))

Answer 16

don't bother with what I said.... for limit x-> infinity it's just sqrt(x^2) / 2x

Answer 17

ok so starting over, i factor whats in the sqrt then factor the bottom, then cancel out whatever i can
\[\frac{ \sqrt{(x-3)(x+3)} }{ 2(x-3) }\]
\[\frac{ \sqrt{(x+3)} }{ 2 }\]

Answer 18

hmm, that's not allowed... but here's an easy way to think about it

Answer 19

imagine putting in x=one million or so...

Answer 20

the -9 becomes insignificant
the -6 becomes insignificant
you're left with rad( 1,000,000^2) (which is just 1,000,000)
divided by 2,000,000

Answer 21

so i can pick any number basically and it will be 1/2

Answer 22

any large number aka x-> infinity

Answer 23

algebraically:

Answer 24

\[\frac{ \sqrt{x^{2}(1-\frac{ 9} { x ^{2} })} }{ x(2-\frac{ 6 }{ x} )}\]

Answer 25

^^ thats the correct way to solve.

Answer 26

factor out x^2 from the term under the radical
factor out x from the term in the denom
(as hartnn was saying before I cut him off)

Answer 27

\[\frac{ x \sqrt{(1-\frac{ 9} { x ^{2} })} }{ x(2-\frac{ 6 }{ x}) }\]

Answer 28

x/x cancels
as x gets large 9/x^2 -> 0
6/x -> 0
all that's left is rad(1) /2 = 1/2

Answer 29

thats an odd way but makes sense lol thanks