need math help. quickly. as in "this is due tommorow". oh, and it's calc 1.

Started by Kitsune Ascendant, September 07, 2006, 11:46:59 PM

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Kitsune Ascendant

the book wants me to find the limit of
(1/(x*(x+1)^(1/2))-1/x, (read: 1 divided by x times the square root of (x+1), minus 1/x) as x goes to 0.

after futzing around with it a bit, I've got it at the point where all I have is
[(x-1)^(1/2)-1]/x. and I can't go any further. I don't see any way to factor an x out of the numerator. and I don't see any way to remove or "change" the x in the denominator to something where the denominator evaluated at 0 is not 0.


any help would be much apreciated.
I may be a bit young to be worrying about it so much, but I'm not changing this sig until I find true love.
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insanekaosx

In a worst case scenario, claim the answer is x-1 and hope for the best :P

No, actually, if I remember correctly, and I probably don't, there should be something about multiplying be a reciprocal or something to get rid of thigns.... >> Math was never my strong point, but I can try ^^;;

Supercheese

Erm, I got -1/2 (negative one-half)...

So, it makes out to be:

1 / (x * sqroot(x+1)) - (1/x)

combine with common denominator:

- sqroot(x+1) + 1 = 0
x * sqroot(x+1)

Multiply both sides by sqroot(x+1)...

(- sqroot(x+1) +1) / x = 0

Wait, limit rules can be done graphing/solving and looking at x approaching 0...

So, not sure why I did all this work, maybe makes it easier to graph/solve >.>  ...

*Ahem* When you *do* solve for x, it's undefined at 0, hence the limit. When x = .0000001, y is really, really close to .5.  When x = -.0000001, y is also really, really close to .5. Isn't that all you need for your proof?

Kitsune Ascendant

Quote from: Supercheese on September 07, 2006, 11:56:46 PM
Erm, I got -1/2 (negative one-half)...

So, it makes out to be:

1 / (x * sqroot(x+1)) - (1/x)

combine with common denominator:

- sqroot(x+1) + 1 = 0
x * sqroot(x+1)

Multiply both sides by sqroot(x+1)...

(- sqroot(x+1) +1) / x = 0

Wait, limit rules can be done graphing/solving and looking at x approaching 0...

So, not sure why I did all this work, maybe makes it easier to graph/solve >.>  ...

*Ahem* When you *do* solve for x, it's undefined at 0, hence the limit. When x = .0000001, y is really, really close to .5.  When x = -.0000001, y is also really, really close to .5. Isn't that all you need for your proof?

for a previous chapter. this one, we're supposed to do it algebraically, so pleace continue as I'm still at a loss.
I may be a bit young to be worrying about it so much, but I'm not changing this sig until I find true love.
yappities by silverfoxr, and are awesome.  Thanks!

Supercheese

Quote from: Kitsune Ascendant on September 07, 2006, 11:59:24 PM
come back! that's the answer!

I'm here...

The definition of limit (I have it saved on my calculator text editor):

L is the limit of f(x) as x approaches c if and only iff L is the one number you can keep f(x) arbitrarily close to just by keeping x close enough to c, but not equal to c.

If that makes sense.

Castle Pokemetroid

What grade are you in? I'm in the 8th grade that looks hard. Or it's the fact that math questions like that makes more sense and looks more better if it's on paper.

I'd help if that didn't look so advance.

Supercheese

So, algebraically, eh? Then you'll need all those deltas and epsilons.

Ok, should be purdy simple (but isn't. blast): solve this equation:

L = limit, e = epsilon

L - e < f(x) < L + e

If you don't know what I'm talking about, then...

"L is the limit of f(x) as x approaches c if and only if for any positive number epsilon, no matter how small, there is a positive number delta such that if x is within delta units of c but not equal to c, then f(x) is within epsilon units of L."

EDIT: Wait, I'm missing something...

Gah, I can't quite remember something critical... it eludes me...


Kitsune Ascendant

HAH! found it!

I was going about it the wrong way. here's what I did:

multiply 1/x by sqrt(x+1)/sqrt(x+1) to get a common denominator, resulting in 1-sqrt(x+1)/(x*sqrt(x+1))

multiply the result above by [1+sqrt(x+1)]/[1+sqrt(x+1)], which, after factoring out an x left me with -1/(sqrt(x+1)+x+1), which is exactly -1/2 when x is equal to 0.
I may be a bit young to be worrying about it so much, but I'm not changing this sig until I find true love.
yappities by silverfoxr, and are awesome.  Thanks!

Supercheese

... that works too. Just as a point of interest, I found this... http://math.hws.edu/javamath/config_applets/EpsilonDelta.html

Sorry if I wasn't much help... =/   I did get the answer right, though.  ;)