Limit of [( x^1/2 ) - x^2] / (1- x^1/2)

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limit of [( x^1/2 ) - x^2] / (1- x^1/2)

Can anyone help me to solve this question?

I have tried all the method but I still cant get the answer.
 

Answers and Replies

  • #2
is x going to infinity or what?
 
  • #3
HallsofIvy
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I think the only difficulty would be if x is going to 1. Standard trick: multiply both numerator and denominator by 1- x1/2, then factor x1/2- x out of the numerator.
 
  • #4
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I get -1 as the answer if limit is x->1
I think you can use L'hopitals rule.

blumfeld0
 
  • #5
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Yes..x is approaching to 1
 
  • #6
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I get the answer of 3 by using hallsofivy. So, which one correct actually?
 
  • #7
VietDao29
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HallsofIvy said:
I think the only difficulty would be if x is going to 1. Standard trick: multiply both numerator and denominator by 1- x1/2, then factor x1/2- x out of the numerator.
Err, I don't really get this... You are actually multiplying both numerator and denominator by a factor (i.e 1 - x1 / 2), which also tends to 0 as x tends to 1.
I thought we should cancel out as many factors that tend to 0 as possible?
Am I missing something?
[tex]\lim_{x \rightarrow 1} \frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \sqrt{x} \ \frac{1 - \sqrt{x ^ 3}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \sqrt{x} \ \frac{1 - (\sqrt{x}) ^ 3}{1 - \sqrt{x}}[/tex]
frozen7 said:
I get the answer of 3 by using hallsofivy. So, which one correct actually?
Yup. That looks good. You can always test your result by using your calculator. Pick up an x that is close to 1 from the left (say 0.99), and from the right (say 1.01). Plug that in the expression:
[tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{0.99} - 0.99 ^ 2}{1 - \sqrt{0.99}} = 2.97002 \approx 3[/tex]
[tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3[/tex].
Are 2.97002, and 3.03002 close to 3 or close to -1?
So you are correct! Congratulations. :smile:
 
Last edited:
  • #8
Hurkyl
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I thought we should cancel out as many factors that tend to 0 as possible?
Sometimes, you have to make an expression more complicated before you can simplify. :smile: This is a common trick, though: you multiply something by 1 in a clever way that simplifies the terms involved.

I think HOI meant to use [itex]1 + \sqrt{x}[/itex], though. (to rationalize one of the terms)


You could also do this with a change of variable, though they don't seem to teach how to do those in elementary calc classes. :frown: But you can still do it the "hard" way by writing the whole thing as a function of [itex]\sqrt{x}[/itex].
 
  • #9
BobG
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blumfeld0 said:
I get -1 as the answer if limit is x->1
I think you can use L'hopitals rule.
blumfeld0
L'Hopital's rule is the easiest way to solve this, but you get 3 with L'hopital's rule, as well. (1/2-2)/(-1/2)
 
  • #10
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Opps my bad. you're all right it is 3. i dont know how i got -1.
sorry for any confusion

blumfeld0
 
  • #11
more basic way is

[tex]\frac {x^{1/2}-x^2} {1-x^{1/2}} = \sqrt{x} \frac {(1 - \sqrt{x}^3) } { 1 - \sqrt{x}} = \sqrt{x} (1+\sqrt{x} + \sqrt{x}^2)[/tex]

Isn't it clear? no law or rules...

if you remember [tex]1-y^n = (1+y+y^2+ \cdots +y^{n-1}) (1-y)[/tex]
or simply
[tex]1-y^3 = (1+y+y^2)(1-y)[/tex]

"back to basic" is often useful :smile:
 
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  • #12
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Thanks...calculus is exciting..!!
By the way, is [tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3[/tex]
a general answer checking method?
 
  • #13
VietDao29
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maverick6664 said:
more basic way is

[tex]\frac {x^{1/2}-x^2} {1-x^{1/2}} = \sqrt{x} \frac {(1 - \sqrt{x}^3) } { 1 - \sqrt{x}} = \sqrt{x} (1+\sqrt{x} + \sqrt{x}^2)[/tex]

Isn't it clear? no law or rules...

if you remember [tex]1-y^n = (1+y+y^2+ \cdots +y^{n-1}) (1-y)[/tex]
or simply
[tex]1-y^3 = (1+y+y^2)(1-y)[/tex]

"back to basic" is often useful :smile:
Ooops... This approach looks familiar. Have you had a look at post #7??
Hurkyl said:
Sometimes, you have to make an expression more complicated before you can simplify.
Hmm... :tongue2:
Hurkyl said:
But you can still do it the "hard" way by writing the whole thing as a function of [itex]\sqrt{x}[/itex]..
And so, my approach is classified as a hard way? :cry: :cry: :cry:
frozen7 said:
By the way, is [tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3[/tex] a general answer checking method?
In case you just have your calculator with you, then yes, it is. I often use it to check whether my answer is correct or not.
It's based on the idea that when you choose an x near 1, the expression will evaluate to some value very close to 3 (since x approaches 1, the expression will approach 3).
But, watch out for rounding error. For example:
[tex]\lim_{x \rightarrow 0} \frac{1}{\sin ^ 2 x} - \frac{1}{x ^ 2} = \frac{1}{3}[/tex]
If you choose x = 0.01, plug that in your calculator (in radian mode), you'll have:
[tex]\frac{1}{\sin ^ 2 0.01} - \frac{1}{0.01 ^ 2} = 0.33334 \approx \frac{1}{3}[/tex]
But if you choose x = 0.00001:
[tex]\frac{1}{\sin ^ 2 0.00001} - \frac{1}{0.00001 ^ 2} \approx 0[/tex]
So choose the x wisely so that it's close enough, but not too close to 1 (or 0, or whatever value x tends to.), so the calculator can evaluate the expression with the smallest rounding error.
 
  • #14
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frozen7 said:
limit of [( x^1/2 ) - x^2] / (1- x^1/2)
Can anyone help me to solve this question?
I have tried all the method but I still cant get the answer.

It is obvious the numerator overwhelm the denominator.
 
  • #15
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Thanks a lot buddy...:)
 
  • #16
HallsofIvy
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kant said:
It is obvious the numerator overwhelm the denominator.

What exactly do you mean by that?
 

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