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Math Help - Calculating the Limit of a Function!

  1. #1
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    Calculating the Limit of a Function!

    Can someone show me the steps of calculating the limit of the following function?

    When x approaches 1 from the left (1 minus), the limit of exp {(4-5x)/(1-x)}.

    Thank you very much
    Last edited by mr fantastic; October 4th 2010 at 11:53 AM. Reason: Edited title.
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  2. #2
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    Quote Originally Posted by Real9999 View Post
    Can someone show me the steps of calculating the limit of the following function?

    When x approaches 1 from the left (1 minus), the limit of exp {(4-5x)/(1-x)}.

    Thank you very much
    limit of Exp[(4-5x)/(1-x)] as x approaches 1 - Wolfram|Alpha
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  3. #3
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    Oh sorry about the title and thanks for your help.

    But can you actually show me how to come to the 0? I know the answer is 0 but I am struggling in getting 0
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  4. #4
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    Oh my gosh! The Mathematica is awesome!!!
    Last edited by mr fantastic; October 4th 2010 at 04:59 PM.
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  5. #5
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    thank you for letting me know the mathematica!! wow! I am getting very excited now XD
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  6. #6
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    Quote Originally Posted by Real9999 View Post
    Oh sorry about the title and thanks for your help.

    But can you actually show me how to come to the 0? I know the answer is 0 but I am struggling in getting 0

    \displaystyle\frac{4-5x}{1-x}}

    If x<1 and approaching 1, then the denominator is positive and approaching 0.

    The numerator is approaching -1.

    What then is the fraction approaching?
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  7. #7
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    Ok! I have just purchased a copy! and downloading now! thank you so much!!!
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  8. #8
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    Oh, yes... what I was concerning was that since it is 1 - x in the denominator, I cannot directly apply the limit to the fraction.

    But thanks Archie! I know how to work it out now! What I need is to take the inverse, (1/x) / (4 - 5x) and evaluate the limit when x approaches 1 from the left and then take the inverse again. That is, the inverse of the limit of the inverse of a fraction is equivalent to the limit of the original function!

    I am so happy today! Thanks for letting me know the existence of mathematica!
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\frac{4-5x}{1-x}}

    If x<1 and approaching 1, then the denominator is positive and approaching 0.

    The numerator is approaching -1.

    What then is the fraction approaching?
    What you are saying here is to directly apply limit to the denominator? does that actually work?

    I think my approach of taking inverse of the function is still flawed ...
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  10. #10
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    Another thing concerning me is that is e^(4-5x)/(1-x) actually has a limit at or approaches 0? I know that exponential function is always positive and not defined at 0
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    I think the answer giving by the mathematica is wrong... just simply by looking at the graph generated by mathematica, it does not look like an exponential function at all...>_< Please correct me if I am wrong... thank you very much.
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  12. #12
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    \displaystyle\frac{4-5x}{1-x}

    The fraction is not defined for x=1 as this will cause the denominator to be zero.

    x=1 does not cause a problem for the numerator, however.

    As x approaches 1 from below, the fraction is negative, the denominator approaches zero and hence the fraction approaches -\infty

    As x approaches 1 from above, the fraction is positive, the denominator approaches zero and hence the fraction approaches +\infty

    \displaystyle\lim_{y\rightarrow\infty}e^y doesnt exist since e^y\rightarrow\infty as y\rightarrow\infty

    \displaystyle\lim_{y\rightarrow\ -\infty}e^y=0
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  13. #13
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\frac{4-5x}{1-x}

    The fraction is not defined for x=1 as this will cause the denominator to be zero.

    x=1 does not cause a problem for the numerator, however.

    As x approaches 1 from below, the fraction is negative, the denominator approaches zero and hence the fraction approaches -\infty

    As x approaches 1 from above, the fraction is positive, the denominator approaches zero and hence the fraction approaches +\infty

    \displaystyle\lim_{y\rightarrow\infty}e^y doesnt exist since e^y\rightarrow\infty as y\rightarrow\infty

    \displaystyle\lim_{y\rightarrow\ -\infty}e^y=0
    Yes, I see your logic here and perhaps I have complicated the matter Thanks
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