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calculate, if existent, via l'Hopital rule:
lim>3 x^2+x12/x^3x^26x
 one year ago
 one year ago
calculate, if existent, via l'Hopital rule: lim>3 x^2+x12/x^3x^26x
 one year ago
 one year ago

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infosharkBest ResponseYou've already chosen the best response.0
what do you get when u sub in 3 ?
 one year ago

calculus130Best ResponseYou've already chosen the best response.0
i am clueless totally..
 one year ago

infosharkBest ResponseYou've already chosen the best response.0
if you put 3 instead of x, what do you get ?
 one year ago

infosharkBest ResponseYou've already chosen the best response.0
now u can use l'hopital's rule. do the derivative of the top over the derivative of the bottom
 one year ago

calculus130Best ResponseYou've already chosen the best response.0
i have done it that way too but i need step by step to check my answer
 one year ago

infosharkBest ResponseYou've already chosen the best response.0
whats the derivative of the numerator over the derivative of the denom that you got ?
 one year ago

calculus130Best ResponseYou've already chosen the best response.0
if i plug 3 for x then the final answer is 7/9 is that right?
 one year ago

richywBest ResponseYou've already chosen the best response.0
\[\lim_{x\rightarrow 3} \frac{ x^2+x12}{x^3x^26x}\]Plug in and see we have an indeterminate form.\[\frac{ 3^2+312}{3^33^26(3)}=\frac{0}{0}\]Appling L'Hopitals rule\[\lim_{x\rightarrow 3} \frac{ \frac{d}{dx}(x^2+x12)}{\frac{d}{dx}(x^3x^26x)}\]\[\lim_{x\rightarrow 3} \frac{ 2x+1}{3x^22x6}\]\[\frac{2(3)+1}{3(3)^22(3)6}\]\[\frac{6+1}{2766}=\frac{7}{15}\]
 one year ago

calculus130Best ResponseYou've already chosen the best response.0
thanks you so much my friend
 one year ago

richywBest ResponseYou've already chosen the best response.0
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%2Bx12%29%2F%28x%5E3x%5E26x%29
 one year ago

calculus130Best ResponseYou've already chosen the best response.0
thank you again for the link
 one year ago
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