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calculus130

  • 3 years ago

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

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  1. infoshark
    • 3 years ago
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    what do you get when u sub in 3 ?

  2. calculus130
    • 3 years ago
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    i am clueless totally..

  3. infoshark
    • 3 years ago
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    if you put 3 instead of x, what do you get ?

  4. calculus130
    • 3 years ago
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    12/0

  5. calculus130
    • 3 years ago
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    i get 12/0

  6. infoshark
    • 3 years ago
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    try one more time.

  7. calculus130
    • 3 years ago
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    oh yeah 0/0

  8. infoshark
    • 3 years ago
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    now u can use l'hopital's rule. do the derivative of the top over the derivative of the bottom

  9. calculus130
    • 3 years ago
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    i have done it that way too but i need step by step to check my answer

  10. infoshark
    • 3 years ago
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    whats the derivative of the numerator over the derivative of the denom that you got ?

  11. calculus130
    • 3 years ago
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    2/6x-2

  12. infoshark
    • 3 years ago
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    2x+1 / 3x^2-2x-6

  13. infoshark
    • 3 years ago
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    plug in 3 for x.

  14. calculus130
    • 3 years ago
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    if i plug 3 for x then the final answer is 7/9 is that right?

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

  16. calculus130
    • 3 years ago
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    thanks you so much my friend

  17. richyw
    • 3 years ago
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    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

  18. calculus130
    • 3 years ago
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    thank you again for the link

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