# Pls help:math question

1. Dec 29, 2005

### teng125

lim x to 9 [(x^2]-81)/[3-(3^1/2)]??

thanx.......

2. Dec 29, 2005

### siddharth

Can you show what you have done till now with this problem?

3. Dec 29, 2005

### teng125

i have tried to do this question n got the answer of -108.is it correct??

4. Dec 29, 2005

### siddharth

No, it is not right.
If you show exactly how you got your answer of -108 (ie, post all the working and calculations you did) it will be easier for us to help you and show where you went wrong.

5. Dec 29, 2005

### TD

That answer isn't correct, unless you stated the wrong problem. I *think* you may mean

$$\mathop {\lim }\limits_{x \to 9} \frac{{x^2 - 81}}{{3 - \sqrt x }}$$

6. Dec 29, 2005

### teng125

ya,that is my question.thank you very much.....

7. Dec 29, 2005

### TD

No problem, but be careful when "writing math" ;)

8. Dec 29, 2005

### fargoth

yeah, try using latex... just click on the equation TD wrote and look at the code, its worth learning.

9. Dec 30, 2005

### HallsofIvy

Okay, the problem is
$$\mathop {\lim }\limits_{x \to 9} \frac{{x^2 - 81}}{{3 - \sqrt x }}$$

do you now see how to get the limit?

The problem is that if you just replace x by 9, both numerator and denominator are 0. (If the denominator were not 0, just calculate the value at x= 9. If the denominator is 0 but the numerator not, there is no limit.)

But since both become 0 at x= 9, that means we can factor! x2- 81 is a "difference of squares" so x2- 81= (x- 9)(x+ 9) and then we can treat x- 9 as a "difference of squares also: $x- 9= (\sqrt(x))^2- 3^2= (\sqrt{x}- 3)(\sqrt{x}+ 3)$
Can you finish it from there?

10. Dec 30, 2005

### fargoth

well i think he did even before posting the question, because he got a right answer, and just asked if its right.

11. Dec 30, 2005

### HallsofIvy

Yeah, I just reread the original post and that is implied.