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Math Help - use pricise defination (limit) to prove

  1. #1
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    use pricise defination (limit) to prove

    lim x/x-1 = 1

    x → infinity

    , then

    got to use this

    so

    x/x-1 = 1

    x/x-1 -1 = x/x-1 - x-1/x-1

    = x-(x-1)/x-1

    so 1/x-1 < E

    am i on the right track? thats as far as i got ,

    some1 help me solve this

    cheers
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  2. #2
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    Re: use pricise defination (limit) to prove

    Quote Originally Posted by arsenal12345 View Post
    lim x/x-1 = 1

    x → infinity

    , then

    got to use this

    so

    x/x-1 = 1

    x/x-1 -1 = x/x-1 - x-1/x-1

    = x-(x-1)/x-1

    so 1/x-1 < E

    am i on the right track? thats as far as i got ,

    some1 help me solve this

    cheers
    To prove that \displaystyle \begin{align*} \lim_{x \to \infty} \frac{x}{x - 1} = 1 \end{align*}, you need to show that \displaystyle \begin{align*} x > N \implies \left|\frac{x}{x - 1} - 1 \right| < \epsilon \end{align*}

    \displaystyle \begin{align*} \left|\frac{x}{x - 1} - 1\right| &< \epsilon \\ \left|\frac{x}{x - 1} - \frac{x - 1}{x - 1}\right| &< \epsilon \\ \left|\frac{1}{x - 1}\right| &< \epsilon \\ \frac{|1|}{|x - 1|} &< \epsilon \\ \frac{1}{|x - 1|} &< \epsilon \\ |x - 1| &> \epsilon \\ x - 1 &> \epsilon \textrm{ since we are making }x \textrm{ be some very large positive value} \\ x &> 1 + \epsilon \end{align*}

    So by letting \displaystyle \begin{align*} N = 1 + \epsilon \end{align*} you will be able to prove \displaystyle \begin{align*} x > N \implies \left|\frac{x}{x - 1} - 1\right| < \epsilon \end{align*}. See how you go
    Thanks from arsenal12345
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  3. #3
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    Re: use pricise defination (limit) to prove

    Thankyou thankyou thankyou
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  4. #4
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    Re: use pricise defination (limit) to prove

    umm what is the purpose of N ,? what does it represent ?
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  5. #5
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    Re: use pricise defination (limit) to prove

    In the case where \displaystyle \begin{align*} c \end{align*} is a number, then you would prove \displaystyle \begin{align*} \lim_{x \to c}f(x) = L \end{align*} by showing that \displaystyle \begin{align*} |x - c| < \delta \implies |f(x) - L| < \epsilon \end{align*}. What this translates to is that as you make \displaystyle \begin{align*} x \end{align*} approach \displaystyle \begin{align*} c \end{align*} and reduce the distance between them, then \displaystyle \begin{align*} f(x) \end{align*} approaches \displaystyle \begin{align*} L \end{align*} and the distance between them also gets smaller.

    But when you want to make \displaystyle \begin{align*} x \to \infty \end{align*}, you can't make \displaystyle \begin{align*} x \end{align*} "close in" on a value. So instead we say that no matter how big we make \displaystyle \begin{align*} x \end{align*}, we will still have \displaystyle \begin{align*} f(x) \end{align*} close in on \displaystyle \begin{align*} L \end{align*}. To do this, provided we choose some very large value for \displaystyle \begin{align*} x \end{align*}, say \displaystyle \begin{align*} N \end{align*}, if we can show that any value greater than this will still make \displaystyle \begin{align*} f(x) \end{align*} approach \displaystyle \begin{align*} L \end{align*}, then we prove \displaystyle \begin{align*} \lim_{x \to \infty} f(x) = L \end{align*}.
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  6. #6
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    Re: use pricise defination (limit) to prove

    ah now i get what u said in the answear, many thanks brother.
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