Limit of (x²−81)/(3−√3) as x approaches 9

[teng125]

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

thanx...

 

[siddharth]

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

 

[teng125]

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

 

[siddharth]

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

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.

 

[TD]

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

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

In that case, your answer is correct

 

[teng125]

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

 

[TD]

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

 

[fargoth]

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

 

[HallsofIvy]

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

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?

 

[fargoth]

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

 

[HallsofIvy]

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