## anonymous 3 years ago calculate, if existent, via l'Hopital rule: lim-->3 x^2+x-12/x^3-x^2-6x

1. anonymous

what do you get when u sub in 3 ?

2. anonymous

i am clueless totally..

3. anonymous

if you put 3 instead of x, what do you get ?

4. anonymous

12/0

5. anonymous

i get 12/0

6. anonymous

try one more time.

7. anonymous

oh yeah 0/0

8. anonymous

now u can use l'hopital's rule. do the derivative of the top over the derivative of the bottom

9. anonymous

i have done it that way too but i need step by step to check my answer

10. anonymous

whats the derivative of the numerator over the derivative of the denom that you got ?

11. anonymous

2/6x-2

12. anonymous

2x+1 / 3x^2-2x-6

13. anonymous

plug in 3 for x.

14. anonymous

if i plug 3 for x then the final answer is 7/9 is that right?

15. richyw

$\lim_{x\rightarrow 3} \frac{ x^2+x-12}{x^3-x^2-6x}$Plug in and see we have an indeterminate form.$\frac{ 3^2+3-12}{3^3-3^2-6(3)}=\frac{0}{0}$Appling L'Hopitals rule$\lim_{x\rightarrow 3} \frac{ \frac{d}{dx}(x^2+x-12)}{\frac{d}{dx}(x^3-x^2-6x)}$$\lim_{x\rightarrow 3} \frac{ 2x+1}{3x^2-2x-6}$$\frac{2(3)+1}{3(3)^2-2(3)-6}$$\frac{6+1}{27-6-6}=\frac{7}{15}$

16. anonymous

thanks you so much my friend

17. richyw

for future reference, if you are just checking your answers, see if wolfram will give you the steps first... http://www.wolframalpha.com/input/?i=lim+x-%3E3+%28x%5E2%2Bx-12%29%2F%28x%5E3-x%5E2-6x%29

18. anonymous

thank you again for the link