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Math Help - Limit of e^x

  1. #1
    Senior Member Paze's Avatar
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    Limit of e^x

    When working with a limit of e^x such as:

    The limit as x approaches infinity of \frac{x^2}{1-e^x}

    Can I simply state that it tends towards 1/((e^inf)/inf) and therefore to 0, or is the limit not yet defined since it tends towards inf/inf below 1? Thanks.
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  2. #2
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    Re: Limit of e^x

    Why do you think that 1/((e^inf)/inf) is 0? In general, you can't compare infinities.
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    Re: Limit of e^x

    Quote Originally Posted by emakarov View Post
    Why do you think that 1/((e^inf)/inf) is 0? In general, you can't compare infinities.
    So I'll need l'hopital right?
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    Re: Limit of e^x

    Yes, you could use L'Hopital's rule. To find limits, you need some database of standard facts, e.g., \frac{a_nx^n+\dots+a_0}{b_nx^n+\dots+b_0}\to\frac{  a_n}{b_n} as x\to\infty provided a_n\ne0 and b_n\ne0. The fact that the \lim_{x\to\infty}\frac{x^n}{a^x}=0 where a>1 is one of those standard facts, but one has to prove it for the first time.
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    Re: Limit of e^x

    You could also use the MacLaurin series for e^x,  1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot so that 1- e^x= -(x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}.

    Now you have \frac{x^2}{1- e^x}= -\frac{x^2}{x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}.

    Divide both numerator and denominator by x^2 to get -\frac{1}{\frac{1}{x}+ \frac{1}{2}+ \frac{x}{6}+ \cdot\cdot\cdot} which goes to infinity as x goes to infinity. Notice that, by this argument, \lim_{x\to\infty} \frac{x}{1- e^x}= 1.
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    Senior Member Paze's Avatar
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    Re: Limit of e^x

    Quote Originally Posted by HallsofIvy View Post
    You could also use the MacLaurin series for e^x,  1+ x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot so that 1- e^x= -(x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}.

    Now you have \frac{x^2}{1- e^x}= -\frac{x^2}{x+ \frac{x^2}{2}+ \frac{x^3}{6}+ \cdot\cdot\cdot}.

    Divide both numerator and denominator by x^2 to get -\frac{1}{\frac{1}{x}+ \frac{1}{2}+ \frac{x}{6}+ \cdot\cdot\cdot} which goes to infinity as x goes to infinity. Notice that, by this argument, \lim_{x\to\infty} \frac{x}{1- e^x}= 1.
    According to calculators and l'hopitals, we have it tending towards 0. Which one is correct...?
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    Re: Limit of e^x

    Quote Originally Posted by Paze View Post
    According to calculators and l'hopitals, we have it tending towards 0. Which one is correct...?
    In both cases the limit is 0.
    I myself find Prof. Ivey's post confusing. But his point is valid: the series approach gives that limit.
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    Re: Limit of e^x

    Quote Originally Posted by Plato View Post
    In both cases the limit is 0.
    I myself find Prof. Ivey's post confusing. But his point is valid: the series approach gives that limit.
    Interesting. I'm going to expand on this idea for my assignment (I hate regurgitating trivial information from a blackboard and I don't care if I get downgraded for veering off course)
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