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Need help with a limit

  1. Jul 24, 2008 #1
    what is the limit:

    lim (1 - c_1 * epsilon) ^ (c_2/epsilon) as epsilon-->0, c_1 and c_2 are const.
  2. jcsd
  3. Jul 24, 2008 #2
    First, I'll LaTeX this so it's easier to read:

    [tex]\lim_{\epsilon \rightarrow 0}{(1 - c_1\epsilon)^{c_2/\epsilon}}[/tex]

    Let y = [tex](1 - c_1\epsilon)^{c_2/\epsilon}[/tex] and consider

    [tex]\ln{y} = c_2/\epsilon \ln{(1 - c_1\epsilon)} = \frac{ \ln{(1 - c_1\epsilon)} }{\epsilon/c_2}}[/tex].

    Note that the numerator and denominator both approach zero as [tex]\epsilon[/tex] tends to zero. Thus we may apply l'Hopital's rule:

    [tex]\lim_{\epsilon \rightarrow 0}{\ln{y}} = \lim_{\epsilon \rightarrow 0}{\frac{\displaystlye{\frac{-c_1}{1-c_1\epsilon}}}{1/c_2}} = \lim_{\epsilon \rightarrow 0}{\frac{-c_1 c_2}{1 - c_2\epsilon}} = -\infty[/tex].

    By the continuity of the exponential function, we thus have that the original limit is equal to

    [tex]\lim_{x \rightarrow -\infty}{\exp{x}} = 0[/tex]

    and we're done.
  4. Jul 24, 2008 #3

    god bless the scots for inventing logarithms
  5. Jul 24, 2008 #4


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    zpconn tried to give you way too much of the solution. We're supposed to give hints here not full solutions. But zpconn also made a mistake. The limit isn't zero. What is it?
  6. Jul 24, 2008 #5
    I apologize for giving too much away. And I did make quite the stupid mistake, didn't I! I apologize for this as well. Hastiness causes such things. But in order to not reveal anything more, I won't say any specifics.
  7. Jul 24, 2008 #6


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    Thanks, zpconn!
  8. Jul 24, 2008 #7
    I was actually able to learn a lot from zpconn's attempted solution and was able to finish myself.

    However, I was never taught this rule in my textbook or my professor, which is saddening. When is this usually introduced in calculus?
  9. Jul 25, 2008 #8
    i'm not sure what mistake you fellas are talking about... is it that zpconn assumed the constants to be positive? that actually WAS an assumption i forgot to mention, so i'm fine with that.

    and what's all that "hints only" policy? this problem came up as a possible shortcut in designing a more efficient algorithm for a certain biology problem. so by not giving me the solution you would only postpone the date cancer will be cured:)
  10. Jul 25, 2008 #9
    Well, I feel that I'm obligated to point out my mistake so I don't mislead you anymore, but I also must abide by the forum rules here and not give away too much. This is an awkward situation to say the least!

    I somehow managed to write down the wrong limit part way through the solution, and that threw off everything else. The general strategy of the rest of the solution is fine, though.

    The correct answer depends on the constants c1 and c2.
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