Question

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

Answer 1

what do you get when u sub in 3 ?

Answer 2

i am clueless totally..

Answer 3

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

Answer 4

12/0

Answer 5

i get 12/0

Answer 6

try one more time.

Answer 7

oh yeah 0/0

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

2/6x-2

Answer 12

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

Answer 13

plug in 3 for x.

Answer 14

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

Answer 15

\[\lim_{x\rightarrow 3} \frac{ x^2+x-12}{x^3-x^2-6x}\]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\[\lim_{x\rightarrow 3} \frac{ \frac{d}{dx}(x^2+x-12)}{\frac{d}{dx}(x^3-x^2-6x)}\]\[\lim_{x\rightarrow 3} \frac{ 2x+1}{3x^2-2x-6}\]\[\frac{2(3)+1}{3(3)^2-2(3)-6}\]\[\frac{6+1}{27-6-6}=\frac{7}{15}\]

Answer 16

thanks you so much my friend

Answer 17

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

Answer 18

thank you again for the link