I'm suppose to evaluate the questions below but i dont really know what to do....the first one results to 0/0 and i tried to factor the down part. I got (x-8)(x^(2)+8x+64) for the denominator and i also got 3(x-1)(x-2) for the numerator. Please i dont know what to do from there... help me, i know this is suppose to be easy questions.

lim x-->2
(3x^(2)-9x+6)/(x^(3)-8)

lim x-->-2
((3x+7)^(-1)-1^(-1))/(x+2)

2. Originally Posted by Ttops
I'm suppose to evaluate the questions below but i dont really know what to do....the first one results to 0/0 and i tried to factor the down part. I got (x-8)(x^(2)+8x+64) for the denominator and i also got 3(x-1)(x-2) for the numerator. Please i dont know what to do from there... help me, i know this is suppose to be easy questions.

lim x-->2
(3x^(2)-9x+6)/(x^(3)-8)

lim x-->-2
((3x+7)^(-1)-1^(-1))/(x+2)
L'Hopitals Rule

3. what does that mean?

4. Originally Posted by Ttops
I'm suppose to evaluate the questions below but i dont really know what to do....the first one results to 0/0 and i tried to factor the down part. I got (x-8)(x^(2)+8x+64) for the denominator and i also got 3(x-1)(x-2) for the numerator. Please i dont know what to do from there... help me, i know this is suppose to be easy questions.

lim x-->2
(3x^(2)-9x+6)/(x^(3)-8)

As both polynomials have 2 as one of their roots, you can factor out $\displaystyle x-2$ from both, so...

lim x-->-2
((3x+7)^(-1)-1^(-1))/(x+2)

I can't understand this expression (in particular that extremely weird (-1)^(-1)...?!).

It'd be a good idea if you read the Help LaTeX section to write mathematics.

Tonio

5. Originally Posted by Ttops
I'm suppose to evaluate the questions below but i dont really know what to do....the first one results to 0/0 and i tried to factor the down part. I got (x-8)(x^(2)+8x+64) for the denominator and i also got 3(x-1)(x-2) for the numerator. Please i dont know what to do from there... help me, i know this is suppose to be easy questions.

lim x-->2
(3x^(2)-9x+6)/(x^(3)-8)

lim x-->-2
((3x+7)^(-1)-1^(-1))/(x+2)
For the first one L'hospital's rule is not needed note that

$\displaystyle \displaystyle \lim_{x \to 2}\frac{3x^2-9x+6}{x^3-8}=\lim_{x \to 2}\frac{3(x-2)(x-1)}{(x-2)(x^2+2x+4)}=\lim_{x \to 2}\frac{3(x-1)}{(x^2+2x+4)}$

As for the 2nd one I can't make it out does it say

$\displaystyle \displaystyle \lim_{x \to -2}\frac{(3x+7)^{-1}-1^{-1}}{(x+2)} = \lim_{x \to -2}\frac{\frac{1}{(3x+7)}-1}{(x+2)}=% \lim_{x \to -2}\frac{1-(3x+7)}{(3x+7)(x+2)}=\lim_{x \to -2}\frac{-3x-6}{(3x+7)(x+2)}=\lim_{x \to -2}\frac{-3}{(3x+7)}$

Edit: it works out fine

6. Thanks alot ... but the second answer was wrong thou.

7. Originally Posted by tonio
I can't understand this expression (in particular that extremely weird (-1)^(-1)...?!).

It'd be a good idea if you read the Help LaTeX section to write mathematics.

Tonio
I've posted a screen shot of the question

8. If you missed the 2nd one, there was an obvious typo in the denominator. That also means you didn't work though the problem yourself and Now this looks like it is part of an assignment. .

9. Originally Posted by TheEmptySet
If you missed the 2nd one, there was an obvious typo in the denominator. That also means you didn't work though the problem yourself and Now this looks like it is part of an assignment. .

(Forum rule #6: http://www.mathhelpforum.com/math-he...hp?do=vsarules)