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

1. Dec 27, 2005

### 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 cant get the answer.

2. Dec 27, 2005

### math-chick_41

is x going to infinity or what?

3. Dec 27, 2005

### HallsofIvy

Staff Emeritus
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. Dec 27, 2005

### blumfeld0

I get -1 as the answer if limit is x->1
I think you can use L'hopitals rule.

blumfeld0

5. Dec 27, 2005

### frozen7

Yes..x is approaching to 1

6. Dec 27, 2005

### frozen7

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

7. Dec 27, 2005

### VietDao29

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?
$$\lim_{x \rightarrow 1} \frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \sqrt{x} \ \frac{1 - \sqrt{x ^ 3}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \sqrt{x} \ \frac{1 - (\sqrt{x}) ^ 3}{1 - \sqrt{x}}$$
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.

Last edited: Dec 27, 2005
8. Dec 27, 2005

### Hurkyl

Staff Emeritus
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 $\sqrt{x}$.

9. Dec 27, 2005

### BobG

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. Dec 27, 2005

### blumfeld0

Opps my bad. you're all right it is 3. i dont know how i got -1.
sorry for any confusion

blumfeld0

11. Dec 27, 2005

### 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

Last edited: Dec 28, 2005
12. Dec 27, 2005

### 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?

13. Dec 28, 2005

### VietDao29

Ooops... This approach looks familiar. Have you had a look at post #7??
Hmm... :tongue2:
And so, my approach is classified as a hard way?
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.

14. Dec 28, 2005

### kant

It is obvious the numerator overwhelm the denominator.

15. Dec 28, 2005

### frozen7

Thanks a lot buddy...:)

16. Dec 28, 2005

### HallsofIvy

Staff Emeritus
What exactly do you mean by that?