anonymous
  • anonymous
calculate, if existent, via l'Hopital rule: lim-->3 x^2+x-12/x^3-x^2-6x
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
what do you get when u sub in 3 ?
anonymous
  • anonymous
i am clueless totally..
anonymous
  • anonymous
if you put 3 instead of x, what do you get ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
12/0
anonymous
  • anonymous
i get 12/0
anonymous
  • anonymous
try one more time.
anonymous
  • anonymous
oh yeah 0/0
anonymous
  • anonymous
now u can use l'hopital's rule. do the derivative of the top over the derivative of the bottom
anonymous
  • anonymous
i have done it that way too but i need step by step to check my answer
anonymous
  • anonymous
whats the derivative of the numerator over the derivative of the denom that you got ?
anonymous
  • anonymous
2/6x-2
anonymous
  • anonymous
2x+1 / 3x^2-2x-6
anonymous
  • anonymous
plug in 3 for x.
anonymous
  • anonymous
if i plug 3 for x then the final answer is 7/9 is that right?
richyw
  • 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\[\lim_{x\rightarrow 3} \frac{ \frac{d}{dx}(x^2+x-12)}{\frac{d}{dx}(x^3-x^2-6x)}\]\[\lim_{x\rightarrow 3} \frac{ 2x+1}{3x^2-2x-6}\]\[\frac{2(3)+1}{3(3)^2-2(3)-6}\]\[\frac{6+1}{27-6-6}=\frac{7}{15}\]
anonymous
  • anonymous
thanks you so much my friend
richyw
  • 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
anonymous
  • anonymous
thank you again for the link

Looking for something else?

Not the answer you are looking for? Search for more explanations.