Question

I need help rationalizing the numerator of:
(1 + cx^2-sqrt(1+x^2))/x^4
The intent is to evaluate the limit as x approaches 0, & to find all values of c for which the limit exists

Answer 1

\[\frac{1 + cx^2-\sqrt{1+x^2}}{x^4}\] like that?

Answer 2

yes

Answer 3

\[\frac{(1 + cx^2)-\sqrt{1+x^2}}{x^4}\times \frac{(1 + cx^2)+\sqrt{1+x^2}}{(1 + cx^2)+\sqrt{1+x^2}}\]

Answer 4

\[\frac{(1+cx^2)^2-(1+x^2)}{x^4\left (1 + cx^2)+\sqrt{1+x^2}\right)}\]

Answer 5

multiply out, leave the denominator factored, get
\[\frac{1+2cx^2+c^2x^4-1-x^4}{x^4\left ((1 + cx^2)+\sqrt{1+x^2}\right)}\]
\[=\frac{2cx^2+(c^2-1)x^4}{x^4\left ((1 + cx^2)+\sqrt{1+x^2}\right)}\]

Answer 6

i guess you can factor a \(x^2\) from the numerator and cancel, not sure what else you can do

Answer 7

there's no way to get rid of \[\sqrt{1+x^2}\]

Answer 8

without undoing what we've just done?
rationalizing the numerator was the Professor's hint on how to evaluate the limit, but this looks counterproductive

Answer 9

you can get rid of it in the numerator, but it shows up in the denominator

Answer 10

that's what I thought, that's why is seems pointless. Thanks for helping me out, I had the algebra a little messed up

Answer 11

i tried playing with a couple different values of \(c\) you can try as well
1 does not work, .1 does not, but .5 does
http://www.wolframalpha.com/input/?i=%281+%2B.5x^2-sqrt%281%2Bx^2%29%29%2Fx^4
lets see if there is more algebra to be done

Answer 12

it appears c must be <1

Answer 13

ok a little help from worlfram shows me that the only \(c\) that works is \(\frac{1}{2}\)

Answer 14

I see that.

Answer 15

so maybe i messed up with the algebra, because i think the numerator should therefore be \(2c-1)+\text{blah}x^2\)

Answer 16

I think the algebra is correct.

Answer 17

if you use .01 for x, and .5 for c, it shows the limit approaching .12

Answer 18

ok in the numerator we get
\[(2c-1)x^2+c^2x^4\] so at least there is the \(2c-1\) term we need

Answer 19

wait, which step did you correct?

Answer 20

yes, it is \(\frac{1}{8}\)
http://www.wolframalpha.com/input/?i=limit+x+to+0+%281+%2B.5x^2-sqrt%281%2Bx^2%29%29%2Fx^4
look at the series expansion on the bottom here
you will see why it only works if \(c=\frac{1}{2}\)

http://www.wolframalpha.com/input/?i=limit+x+to+0+%281+%2Bcx^2-sqrt%281%2Bx^2%29%29%2Fx^4

Answer 21

ok here is my mistake
\[\frac{1+2cx^2+c^2x^4-1-x^4}{x^4\left ((1 + cx^2)+\sqrt{1+x^2}\right)}\] should be
\[\frac{1+2cx^2+c^2x^4-1-x^2}{x^4\left ((1 + cx^2)+\sqrt{1+x^2}\right)}\]

Answer 22

numerator is
\[\frac{(2c-1)x^2+c^2x^4}{x^4\left ((1 + cx^2)+\sqrt{1+x^2}\right)}\]

Answer 23

now i can see it, i made a silly mistake before

Answer 24

\(2c-1\) must be zero, otherwise you have an \(x^2\) term in the numerator which will mess everything up
sorry about the algebra goof, i was bush league mistake

Answer 25

it's ok, I was struggling with this one really bad, I really appreciate your help

Answer 26

no problem