How can I solve this limit algebraically without using the conjugate method?

[Schrodinger's Cat]

I'm trying to solve this limit algebraically

limit as x approaches 1 (x^.5-x^2)/(1-x^.5).

I tried multiplying by the conjugate 1+x^.5 but that left the denominator with 1-x and I was unable to cancel that out.

Can anyone help me out?

 

[Hurkyl]

Well, can you show what you got when you tried that?

 

[Schrodinger's Cat]

Sure.

I got (x^.5+x-x^2-x^(5/2))/(1-x).

I can't figure out how to factor the top so as to cancel out the 1-x part.

 

[Hurkyl]

Well, you can always do long-division to see if there's anything left-over. Or, which amounts to the same thing, rewrite your numerator with tricks like:

-x^2.5 = (1-x) x^1.5 - x^1.5


(Actually, you could have done long division right from the beginning, without bothering with the conjugate -- I wonder why this isn't often taught?)


Hrm. Maybe the person who wrote the problem expected you to split the numerator into parts that do and do not have fractional powers of x. i.e. as:

x^.5 (1 - x^2) + (x - x^2)

I'm not sure if I prefer this, or the division method I mentioned.


p.s. you meant x^(5/2) not x^5/2.

 

[Schrodinger's Cat]

I tried getting it into other formats such as x^.5-x^2/1-x^.5 but it didn't work out too well.

 

[Schrodinger's Cat]

Well, you can always do long-division to see if there's anything left-over.

How exactly would that work out (I'm kinda lost here)? I'm just trying to rationalize the function so as to find the limit for it.

 

[Hurkyl]

(I've added a bit more to my previous post)


Well, division here works just like ordinary division

      x^1.5 + x
    ------------------------
1-x | x^.5 + x - x^2 - x^2.5
      x^1.5 - x^2.5
      ----------------------
      x^.5 + x - x^1.5 - x^2
      x - x^2
      ----------------------
      x^.5 - x^1.5

I've done the first two steps for you -- first, I chose x^1.5, because if I multiply the divisor by x^1.5, I can cancel out the biggest term in the dividend after subtracting.

(I've decided to have high powers of x be the "biggest" term -- you can do the opposite if you prefer)

 

[Schrodinger's Cat]

-x^2.5 = (1-x) x^1.5 - x^1.5

How did you get that?

x^.5 (1 - x^2) + (x - x^2)

Not exactly sure how you got that as well...

Well, division here works just like ordinary division

So, what about the remainder resulting from the long division? You're saying that it'll factor into ((1-x)(x^1.5+x))/(1-x). The 1-x will cancel out leaving x^1.5 +x?

 

[Schrodinger's Cat]

p.s. you meant x^(5/2) not x^5/2.

Yeah, my mistake. Not used to typing it all out on a keyboard.:tongue2:

Edited.

 

[Hurkyl]

How did you get that?

Try multiplying it out! This is a very, very useful trick -- you should learn it!

Not exactly sure how you got that as well...

Again, try multiplying it out!


(Or did you mean "how did I think of it"?)

 

[Schrodinger's Cat]

Wait, I see how you got those two parts though I'm not sure how that would help me out though.:uhh:

 

[wurth_skidder_23]

Starting from the original equation, it can be solved with partial fractions to be:
x^1.5-1+(1-x^1.5)/(1-x^0.5)
=x^1.5-1+(1-x^0.5)*(1+x^0.5+x)/(1-x^0.5)

Limit as x -> 1 is therefore 3.

 

[Hurkyl]

So, what about the remainder resulting from the long division? You're saying that it'll factor into ((1-x)(x^1.5+x))/(1-x). The 1-x will cancel out leaving x^1.5 +x?

If you stop there, then yes. (Except you swapped the quotient and the remainder):

(x^.5 + x - x^2 - x^2.5) / (1-x)
= (x^1.5 + x) + (x^.5 - x^1.5) / (1-x)

but the division isn't finished -- you can get a better remainder!

 

[Schrodinger's Cat]

Again, try multiplying it out!

Yeah, I see now but I don't see how that will help me factor the top though...

 

[Hurkyl]

Wait, I see how you got those two parts though I'm not sure how that would help me out though.

Or, which amounts to the same thing, rewrite your numerator with tricks like:

-x^2.5 = (1-x) x^1.5 - x^1.5


(Actually, you could have done long division right from the beginning, without bothering with the conjugate -- I wonder why this isn't often taught?)

I'm suggesting that, if you wanted to do it this way, to try this trick a few times, and see if you wind up with something useful! (This is what I first did)

Hrm. Maybe the person who wrote the problem expected you to split the numerator into parts that do and do not have fractional powers of x. i.e. as:

x^.5 (1 - x^2) + (x - x^2)

Stare at it for a while. Maybe putting it back into conext will help: you want the limit of

$$\frac{\sqrt{x} (1 - x^2) + (x - x^2)}{1 - x}$$

 

[wurth_skidder_23]

Schrodinger's Cat, you should start with the original equation and do the long division. You'll get the answer that I obtained in my earlier reply.

 

[Schrodinger's Cat]

x^1.5-1+(1-x^1.5)

How did you get that part?

 

[wurth_skidder_23]

I did long division of the original equation.

 

[Schrodinger's Cat]

Stare at it for a while. Maybe putting it back into conext will help: you want the limit of

Ok, but I still don't see how to get rid of the 1-x part.

I'll finish the long division though I can't remember it all that well.

 

[Hurkyl]

What if you were working with:

$$\frac{1 - x^2}{1 - x}$$
?

 

[Schrodinger's Cat]

I did long division of the original equation.

I'm getting something different. :uhh: Ordering the polynomials doesn't matter right?

 

[Schrodinger's Cat]

What if you were working with:

So that should leave the equation at x^1.5(1+x)+x and therefore a limit of 3?

 

[Schrodinger's Cat]

Starting from the original equation, it can be solved with partial fractions to be:
x^1.5-1+(1-x^1.5)/(1-x^0.5)
=x^1.5-1+(1-x^0.5)*(1+x^0.5+x)/(1-x^0.5)

Limit as x -> 1 is therefore 3.

:uhh: How did you get 3?

 

[Hurkyl]

So that should leave the equation at x^1.5(1+x)+x and therefore a limit of 2?

That looks like the right expression. You just plugged "1" in incorrectly.

 

[wurth_skidder_23]

Here is a JPEG image of what I did:

http://us.f2.yahoofs.com/users/415c3ee6z2007f18d/deepsleep_6er/__sr_/71b1scd.jpg?pfAN39EBK6njJ4lZ

 

[Schrodinger's Cat]

That looks like the right expression. You just plugged "1" in incorrectly.

Whoops!:yuck: Ok, got it.

The problem is that I never would've thought of that trick without outside help...

 

[Schrodinger's Cat]

Here is a JPEG image of what I did:

The pics kinda difficult to see but I still don't see how you got 3.

 

[wurth_skidder_23]

The pic should be easier to see now. I got 3 because the final (remainder) term can be factored to be (1+x^0.5+x) which is 3 when x = 1.

 

[Schrodinger's Cat]

I can't even see the pic anymore. So you're supposed to base the limit on the remainder?

 

[wurth_skidder_23]

No, this is a special case. It just so happens that in this case, the x^1.5-1 cancels out when x = 1. The remainder becomes (1-x^1.5)/(1-x^0.5), which can be factored into (1-x^0.5)*(1+x^0.5+x)/(1-x^0.5). This factorization can be obtained by another long division.

 

[daveb]

I can't read the jpeg for some reason, but I also got 3 by decomposing the numerator in the original equation (x^0.5-x^2) = x^0.5(1-(x^0.5)^3) which is a cubic and can be further factored as x^0.5(1-x^0.5)(1+x^0.5+x). The (1-x^0.5) cancel out leaving x^0.5)(1+x^0.5+x). Then take this limit as x goes to 1 gives 3. I imagine that's similar to what wurth did