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The behaviour of e^x near infinity and -infinity

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    I have done an integration and ended up with the result

    [-c/2 * [e^(-2x)]] |^infinity_0 = 1
    The solution is that c=2 so that means to me that e^(2x) must turn into minus 1 for it to equal 1... but I'm not sure.. I've got graphcalc so I've been staring at the graph and I figure that as x goes to infinity that e^x goes to 1... but not sure what to say when x goes to minus infinity?
  2. jcsd
  3. Jun 1, 2008 #2


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    You should remember the following properties of the exponential function:
    • [tex]\lim_{x \to +\infty} e^x = \infty[/tex]
    • [tex]\lim_{x \to -\infty} e^x = 0[/tex]
    • [tex]\lim_{x \to 0} e^x = 1[/tex] (actually, the exponential function is continuous in 0, so one could also just say [tex]e^0 = 1[/tex], which is logical since [itex]x^0 = 1[/itex] for any [itex]x \neq 0[/itex]).
    Last edited: Jun 1, 2008
  4. Jun 1, 2008 #3
    e^x goes to 1 as x goes to 0.
    e^x goes to 0 as x goes to negative infinity
    e^x goes to infinity as x goes to infinity (no limit)

    Is that what you're after?
  5. Jun 1, 2008 #4


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    Then you need a new calculator! e^x does not go anywhere near 1 as x goes to infinity.
    If you must use a calculator, what is e^1000000? What is e^(-100000)?
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