Here's the question you clicked on:
calculus130
calculate, if existent, via l'Hopital rule: lim-->3 x^2+x-12/x^3-x^2-6x
what do you get when u sub in 3 ?
i am clueless totally..
if you put 3 instead of x, what do you get ?
now u can use l'hopital's rule. do the derivative of the top over the derivative of the bottom
i have done it that way too but i need step by step to check my answer
whats the derivative of the numerator over the derivative of the denom that you got ?
if i plug 3 for x then the final answer is 7/9 is that right?
\[\lim_{x\rightarrow 3} \frac{ x^2+x-12}{x^3-x^2-6x}\]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\[\lim_{x\rightarrow 3} \frac{ \frac{d}{dx}(x^2+x-12)}{\frac{d}{dx}(x^3-x^2-6x)}\]\[\lim_{x\rightarrow 3} \frac{ 2x+1}{3x^2-2x-6}\]\[\frac{2(3)+1}{3(3)^2-2(3)-6}\]\[\frac{6+1}{27-6-6}=\frac{7}{15}\]
thanks you so much my friend
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
thank you again for the link