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  1. #1
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    Help with a limit question

    Please determine the following limit if they exist. If the limit doe not exist put DNE.
    lim 2x^3 / x^2 + 10x - 12
    x->infinity.
    Thanks.
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    Quote Originally Posted by rowdy3 View Post
    Please determine the following limit if they exist. If the limit doe not exist put DNE.
    lim 2x^3 / x^2 + 10x - 12
    x->infinity.
    Thanks.
    Do you know L'Hospitals?

    \lim_{x\to\infty} \frac{2x^3}{x^2 + 10x - 12}

    = \lim_{x\to\infty} \frac{6x^2}{2x + 10}

    = \lim_{x\to\infty} \frac{12x}{2} = \lim_{x\to\infty} 6x \to\infty \fbox{DNE}
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  3. #3
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    You may wish to review your "Order of Operations" and add a few more parentheses, but other than that, with rational functions, degree of numerator greater than degree of denominator ==> Definitely DNE.
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    Quote Originally Posted by rowdy3 View Post
    Please determine the following limit if they exist. If the limit doe not exist put DNE.
    lim 2x^3 / x^2 + 10x - 12
    x->infinity.
    Thanks.
    very straight-forward limit ... what do you think?
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  5. #5
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    Quote Originally Posted by rowdy3 View Post
    Please determine the following limit if they exist. If the limit doe not exist put DNE.
    lim 2x^3 / x^2 + 10x - 12
    x->infinity.
    Thanks.
    Hi rowdy3,

    As the numerator contains a higher power of x, then the expression
    increases without bound as x does, hence the limit DNE.

    \frac{x^3}{x^2+10x-12}=\frac{\frac{1}{x}\left(x^3\right)}{\frac{1}{x}  \left(x^2+10x-12\right)}

    =\frac{x^2}{x+10-\frac{12}{x}}

    As x\ \rightarrow\ \infty the term \frac{12}{x}\ \rightarrow\ 0

    \frac{\frac{1}{x}\left(x^2\right)}{\frac{1}{x}(x+1  0)}=\frac{x}{1+\frac{10}{x}}

    As x\ \rightarrow\ \infty the term \frac{10}{x}\ \rightarrow\ 0

    \lim_{x\ \rightarrow\ \infty}\frac{x}{1} DNE

    If you like, do it in one step, eliminating x from the denominator by multiplying by \frac{\left(\frac{1}{x^2}\right)}{\left(\frac{1}{x  ^2}\right)}
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