Question

lim as x approaches infinity > 3x/(sqrt x^2-1)

Answer 1

can you factor out 'x' from the denominator ?

Answer 2

or
can you factor, x^2 from x^2-1 ?

Answer 3

x^2-1 is under radical in the denom. can't remember how to solve this

Answer 4

ok, so does this make sense ?
\((x^2-1)= x^2 (1-1/x^2)\)

Answer 5

yes, it is still under radical. right?

Answer 6

yup.

so now \(\sqrt{x^2 (1-1/x^2)}=....?\)
now can you factor out x ?

Answer 7

i know the answer is 3, but can't get the steps.

Answer 8

does 1/3^2 = 0

Answer 9

\(\sqrt{x^2 (1-1/x^2)}=x\sqrt{1-1/x^2}\)
got that ^ ?

Answer 10

got that

Answer 11

cool, so what gets cancelled ?

Answer 12

x

Answer 13

plug in 3 now?

Answer 14

wait,

x -> infinity!
so, 1/x -> 0

does this make sense ?

Answer 15

no. i have 3/(x-(1/x^2))

Answer 16

i don't know how to use the symbols on my computer for this

Answer 17

but those 'x' got cancelled from numerator and denominator, right ?

so you r left with
\(\dfrac{3}{\sqrt {1-1/x^2}}\)
isn't it ?

Answer 18

yep

Answer 19

so, just plug in 1/x^2 = 0 in that!
what u get ?

Answer 20

why do you plug in 0?

Answer 21

oh, i just said
as x-> infinity
so 1/x -> 0
so, 1/x^2 will also ->0
so we put 1/x^2 = 0 to get the limit :)

Answer 22

thanks

Answer 23

welcome ^_^