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

[frozen7]

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 can't get the answer.

 

[math-chick_41]

is x going to infinity or what?

 

[HallsofIvy]

I think the only difficulty would be if x is going to 1. Standard trick: multiply both numerator and denominator by 1- x^1/2, then factor x^1/2- x out of the numerator.

 

[blumfeld0]

I get -1 as the answer if limit is x->1
I think you can use l'hospital's rule.

blumfeld0

 

[frozen7]

Yes..x is approaching to 1

 

[frozen7]

I get the answer of 3 by using hallsofivy. So, which one correct actually?

 

[VietDao29]

I think the only difficulty would be if x is going to 1. Standard trick: multiply both numerator and denominator by 1- x^1/2, then factor x^1/2- 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 - x^{1 / 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?
\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}}

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:
\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{0.99} - 0.99 ^ 2}{1 - \sqrt{0.99}} = 2.97002 \approx 3
\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3.
Are 2.97002, and 3.03002 close to 3 or close to -1?
So you are correct! Congratulations.

 

[Hurkyl]

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. 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 1 + \sqrt{x}, 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. But you can still do it the "hard" way by writing the whole thing as a function of \sqrt{x}.

 

[BobG]

I get -1 as the answer if limit is x->1
I think you can use l'hospital's 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)

 

[blumfeld0]

Opps my bad. you're all right it is 3. i don't know how i got -1.
sorry for any confusion

blumfeld0

 

[maverick6664]

more basic way is

\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)

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

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

"back to basic" is often useful

 

[frozen7]

Thanks...calculus is exciting..!
By the way, is \frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3
a general answer checking method?

 

[VietDao29]

more basic way is

\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)

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

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

"back to basic" is often useful

Ooops... This approach looks familiar. Have you had a look at post #7??

Sometimes, you have to make an expression more complicated before you can simplify.

Hmm...

But you can still do it the "hard" way by writing the whole thing as a function of \sqrt{x}..

And so, my approach is classified as a hard way?

By the way, is \frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3 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:
\lim_{x \rightarrow 0} \frac{1}{\sin ^ 2 x} - \frac{1}{x ^ 2} = \frac{1}{3}
If you choose x = 0.01, plug that in your calculator (in radian mode), you'll have:
\frac{1}{\sin ^ 2 0.01} - \frac{1}{0.01 ^ 2} = 0.33334 \approx \frac{1}{3}
But if you choose x = 0.00001:
\frac{1}{\sin ^ 2 0.00001} - \frac{1}{0.00001 ^ 2} \approx 0
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.

 

[kant]

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 can't get the answer.

It is obvious the numerator overwhelm the denominator.

 

[frozen7]

Thanks a lot buddy...:)

 

[HallsofIvy]

It is obvious the numerator overwhelm the denominator.

What exactly do you mean by that?