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Math Help - need help with an easy limit

  1. #1
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    need help with an easy limit

    lim x->1 (x^3 - 3x +2)/(x^4 - 4x + 3) = ?

    I tried with L'Hospital's rule but it seems it doesn't work (it still gives me 0/0... Maybe I didn't apply it well?). Any suggestions?

    Thanks in advance !

    [Sorry if it's a dumb question, but I am not very good at math...]
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  2. #2
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    Quote Originally Posted by newby View Post
    lim x->1 (x^3 - 3x +2)/(x^4 - 4x + 3) = ?

    I tried with L'Hospital's rule but it seems it doesn't work (it still gives me 0/0... Maybe I didn't apply it well?). Any suggestions?

    Thanks in advance !

    [Sorry if it's a dumb question, but I am not very good at math...]
    \lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3}= \lim_{x \rightarrow 1  } \frac{ 3x^2-3}{4x^3-4 }= \lim_{ x \rightarrow 1}  \frac{3 }{4 }   \left (   \frac{ x^2-1}{x^3-1 }  \right ) = \lim_{ x \rightarrow 1}  \frac{ 3}{ 4}   \left (  \frac{ x+1}{x^2+x+1 }   \right ) = \frac{ 3}{4 }  \left (   \frac{2 }{3 }  \right ) = \frac{ 1}{ 2}
    hope i'm right
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  3. #3
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    Quote Originally Posted by Raoh View Post
    \lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3}= \lim_{x \rightarrow 1  } \frac{ 3x^2-3}{4x^3-4 }= \lim_{ x \rightarrow 1}  \frac{3 }{4 }   \left (   \frac{ x^2-1}{x^3-1 }  \right ) = \lim_{ x \rightarrow 1}  \frac{ 3}{ 4}   \left (  \frac{ x+1}{x^2+x+1 }   \right ) = \frac{ 3}{4 }  \left (   \frac{2 }{3 }  \right ) = \frac{ 1}{ 2}
    hope i'm right
    Thanks a lot! I feel more clever now !
    Thanks again !
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  4. #4
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    Quote Originally Posted by newby View Post
    lim x->1 (x^3 - 3x +2)/(x^4 - 4x + 3) = ?

    I tried with L'Hospital's rule but it seems it doesn't work (it still gives me 0/0... Maybe I didn't apply it well?). Any suggestions?

    Thanks in advance !

    [Sorry if it's a dumb question, but I am not very good at math...]
    Here comes a slightly different way:

    As long as the quotient is \frac00 you can apply de l'Hopital's rule:

    \lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3} = \lim_{x\to 1}\frac{3x^2-3}{4x^3-4} = \lim_{x\to 1}\frac{6x}{12x^2} = \frac6{12} = \frac12
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  5. #5
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    Quote Originally Posted by earboth View Post
    Here comes a slightly different way:

    As long as the quotient is \frac00 you can apply de l'Hopital's rule:

    \lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3} = \lim_{x\to 1}\frac{3x^2-3}{4x^3-4} = \lim_{x\to 1}\frac{6x}{12x^2} = \frac6{12} = \frac12
    Thank you too! I didn't know I could use the rule many times!
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  6. #6
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    Quote Originally Posted by newby View Post
    Thank you too! I didn't know I could use the rule many times!
    neither i .
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