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calculus130
 3 years ago
calculate, if existent, via l'Hopital rule:
lim>3 x^2+x12/x^3x^26x
calculus130
 3 years ago
calculate, if existent, via l'Hopital rule: lim>3 x^2+x12/x^3x^26x

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infoshark
 3 years ago
Best ResponseYou've already chosen the best response.0what do you get when u sub in 3 ?

calculus130
 3 years ago
Best ResponseYou've already chosen the best response.0i am clueless totally..

infoshark
 3 years ago
Best ResponseYou've already chosen the best response.0if you put 3 instead of x, what do you get ?

infoshark
 3 years ago
Best ResponseYou've already chosen the best response.0now u can use l'hopital's rule. do the derivative of the top over the derivative of the bottom

calculus130
 3 years ago
Best ResponseYou've already chosen the best response.0i have done it that way too but i need step by step to check my answer

infoshark
 3 years ago
Best ResponseYou've already chosen the best response.0whats the derivative of the numerator over the derivative of the denom that you got ?

calculus130
 3 years ago
Best ResponseYou've already chosen the best response.0if i plug 3 for x then the final answer is 7/9 is that right?

richyw
 3 years ago
Best 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\[\lim_{x\rightarrow 3} \frac{ \frac{d}{dx}(x^2+x12)}{\frac{d}{dx}(x^3x^26x)}\]\[\lim_{x\rightarrow 3} \frac{ 2x+1}{3x^22x6}\]\[\frac{2(3)+1}{3(3)^22(3)6}\]\[\frac{6+1}{2766}=\frac{7}{15}\]

calculus130
 3 years ago
Best ResponseYou've already chosen the best response.0thanks you so much my friend

richyw
 3 years ago
Best ResponseYou've already chosen the best response.0for 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

calculus130
 3 years ago
Best ResponseYou've already chosen the best response.0thank you again for the link
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