- #1

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Can anyone help me to solve this question?

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

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- Thread starter frozen7
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- #1

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Can anyone help me to solve this question?

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

- #2

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is x going to infinity or what?

- #3

HallsofIvy

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- #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

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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Err, I don't really get this... You are actually multiplying both numerator and denominator by a factor (i.e 1 - xHallsofIvy said:^{1/2}, then factor x^{1/2}- x out of the numerator.

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]

Yup. That looks good. You can always test your result by using yourfrozen7 said:I get the answer of 3 by using hallsofivy. So, which one correct actually?

[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.

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- #8

Hurkyl

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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 thought we should cancel out as many factors that tend to 0 as possible?

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. 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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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 said:I get -1 as the answer if limit is x->1

I think you can use L'hopitals rule.

blumfeld0

- #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

sorry for any confusion

blumfeld0

- #11

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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

[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

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- #12

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

Hmm... :tongue2:Hurkyl said:Sometimes, you have to make an expression more complicated before you can simplify.

And so, my approach is classified as a hard way?Hurkyl said:But you can still do it the "hard" way by writing the whole thing as a function of [itex]\sqrt{x}[/itex]..

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.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?

It's based on the idea that when you choose an

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

[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

- #14

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frozen7 said:

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

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