Recent content by phrox

Thanks a ton, I just find it so weird thinking of x as in sqrt(x^2) and etc. Wish I had time to practice limits instead of getting an assignment as soon as I'm finished the last lol.

I'm guessing you can't do it simply this way? http://s1301.photobucket.com/user/phrox1/media/image001_zps75e88aa9.jpg.html?sort=3&o=0

no I haven't covered L'Hospital's rule, also I can't directly sub x->0 into this because it just won't work, undefined.

So did is my equation for a) H - 1/x = (sqrt(1/x^2)+1) - sqrt(1/x^2) and for b) it says instead, compute the limit as x-> 0, which would make it not work because all the x's are under ones, making them undefined.

http://i1301.photobucket.com/albums/ag115/phrox1/additional_zps10bf2373.png That's the question ^^, So I came up with H = sqrt(D^2 + 1) Then I got H - D = (sqrt(1/x^2)+1) - D Annndd then: H - 1/x = (sqrt(1/x^2)+1) - sqrt(1/x^2) I have no clue where to go from here, any help is appreciated!

nevermind I understand now, thanks

If you say don't multiply the 2 radical expressions out, am I leaving the (sqrt(5-x)-1)(2+sqrt(x)) in the top, then multiplying the top and bottom by sqrt(5-x)+1? So in the end I'd be multiplying 3 different expressions in the top: (sqrt(5-x)-1)(2+sqrt(x))(sqrt(5-x)+1). Here is what I get after...

By rationalize do you mean multiply the conjugate? I don't see how you can rationalize (multiply top and bottom by the root), but you say you only do it to the top and then only do it to the bottom. I don't think you can do that. Also if I multiply the top conjugate and then bottom conjugate it...

By rationalize do you mean multiply the conjugate? I don't see how you can rationalize (multiply top and bottom by the root), but you say you only do it to the top and then only do it to the bottom. I don't think you can do that. Also if I multiply the top conjugate and then bottom conjugate it...

The only thing I see you can do with that is square it I guess?I'm not sure... Subbing in 4 will just make it 0.

When I multiplied by the conjugate I ended up with 4-x on the denominator. Subbing in 4 would make the limit impossible.

just by x? Because if I multiply by (2+sqrt(x)) it just got long and tedius, thought there was an easier way.

Homework Statement limx->4 (sqrt(5-x)-1) / (2-sqrt(x)) NOT ALLOWED TO USE L'HOSPITALS. Homework Equations The Attempt at a Solution I tried using conjugate of both top and bottom and I couldn't get it to work, but maybe I've done it wrong?.. I'm not allowed to use...

Oh.. Not sure how it's done then

Oh well I just assumed that since it's left continuous at 2 going in a positive direction, then if it's going in a negative direction(to the right) then it would have to be -2.