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Math Help - Testing extremities!

  1. #1
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    Post Testing extremities!

    I have,
    y= e^x / x

    Now, I want to test the extremities, so I sub in
    x -> +ve Infinity
    so,
    y -> 0

    Right? However, when I graphed it on the calculator.. when x reached positive infinity, the y is not reaching zero.. it curved up to positive infinity..

    Can anyone explain this? Thanks!
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  2. #2
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    Quote Originally Posted by Modernized View Post
    I have y= e^x / x
    Does the above mean the following?

    . . . . . y\, =\, \frac{e^x}{x}

    Assuming so...

    Quote Originally Posted by Modernized View Post
    I want to test the extremities...
    Does the above means "find the horizontal asymptote" or "find the limit as x gets arbitrarily large" or something similar?

    Assuming so...

    Quote Originally Posted by Modernized View Post
    ...so I sub in
    x -> +ve Infinity
    Does the above mean "So I take the limit of the expression, as x tends toward positive infinity"?

    Assuming so...

    Quote Originally Posted by Modernized View Post
    so,
    y -> 0
    How did you arrive at this limit value, seeing that the exponential numerator grows much more quickly than does the linear denominator?

    Kindly please reply with corrections, confirmations, and / or clarification. Thank you!
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  3. #3
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    Smile

    Yes, all of your assumptions are correct! Sorry for the doggy worded question!

    "How did you arrive at this limit value, seeing that the exponential numerator grows much more quickly than does the linear denominator?"

    y=e^x / x
    as x approached +ve infinity, the numerator will definitely grow faster than the denominator

    But! my teacher told me that as x -> + infinity,
    y will get smaller and smaller, hence approaches 0.. is this correct?
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  4. #4
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    Quote Originally Posted by Modernized View Post
    Yes, all of your assumptions are correct! Sorry for the doggy worded question!

    "How did you arrive at this limit value, seeing that the exponential numerator grows much more quickly than does the linear denominator?"

    y=e^x / x
    as x approached +ve infinity, the numerator will definitely grow faster than the denominator

    But! my teacher told me that as x -> + infinity,
    y will get smaller and smaller, hence approaches 0.. is this correct?
    It's more likely that your teacher told you that as  x \rightarrow {\color{red}-} \infty then y \rightarrow 0. Either that, or the function is actually y = \frac{e^{{\color{red}-} x}}{x}.
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  5. #5
    Senior Member nikhil's Avatar
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    Quote Originally Posted by Modernized View Post
    I have,
    y= e^x / x

    Now, I want to test the extremities, so I sub in
    x -> +ve Infinity
    so,
    y -> 0

    Right? However, when I graphed it on the calculator.. when x reached positive infinity, the y is not reaching zero.. it curved up to positive infinity..

    Can anyone explain this? Thanks!
    just find limit when x approaches to +infinity
    L'Hospital's rule may be used which straightly gives limit to be +infinity
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