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Thread: limit = e

  1. #1
    Member Greengoblin's Avatar
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    limit = e

    Can someone show me how:

    $\displaystyle \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$
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  2. #2
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    Quote Originally Posted by Greengoblin View Post
    Can someone show me how:

    $\displaystyle \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$
    Say that this limit exists and it equals A.

    Then $\displaystyle =\frac{1}{t}\ln(1+t)=\ln(A)$

    The LHS is also $\displaystyle \frac{\ln(1+t)}{t}$ and when evaluating at t=0 you get 0/0, which is indeterminate and L'Hopitals can be used.

    So $\displaystyle \ln(A)=\frac{\frac{1}{1+t}}{1}$

    Try the limit again, and use algebra to solve for A again.

    edit: This is in the precalc section, so forgive me for using calculus.
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  3. #3
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    Quote Originally Posted by Greengoblin View Post
    Can someone show me how:

    $\displaystyle \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$
    Another way is to sub $\displaystyle u=\frac{1}{t}$ so that $\displaystyle u \to \infty~\text{as}~t \to 0$ to get the famous limit:

    $\displaystyle \lim_{u \to \infty} \left(1+\frac{1}{u}\right)^u = e$

    @Jameson: I am sure that Greengoblin's level is beyond precal.
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  4. #4
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    Yes that would appear so after looking some more
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  5. #5
    Member Greengoblin's Avatar
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    Thank you both so much, I understand it now.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Alternate solution:

    recall that $\displaystyle e^{\ln X} = X$ for any $\displaystyle X > 0$.

    so that:

    $\displaystyle \lim_{t \to 0}(1 + t)^{\frac 1t} = \lim_{t \to 0}\text{exp} \left( \frac 1t \ln (1 + t) \right)$

    $\displaystyle = \text{exp} \left( \lim_{t \to 0} \frac {\ln (1 + t)}t \right)$

    now since $\displaystyle \lim_{t \to 0} \frac {\ln (1 + t)}t = 1$ (By L'Hopital's rule, or power series or ...)

    we have that the limit is $\displaystyle \text{exp}(1) = e$


    This is very similar to Jameson's method, both of these methods you should consider when taking the limit of an expression with the variable in the power.

    That is, let $\displaystyle A $ and $\displaystyle B$ be expressions in a single variable in which the variable appears, then we can usually find various limits that give indeterminate forms by doing the following:

    The method Jameson used:

    Say $\displaystyle y = A^B$

    Then $\displaystyle \ln y = B \ln A$

    so that $\displaystyle \lim \ln y = \lim B \ln A$

    and so $\displaystyle \lim y = e^{\lim B \ln A}$ by taking the anti-log of both sides.


    The method I used:

    simply write $\displaystyle y = A^B$ as $\displaystyle e^{\ln A^B} = e^{B \ln A}$, and take the limit of both sides.

    note that $\displaystyle \lim e^X = e^{\lim X}$
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  7. #7
    Member Greengoblin's Avatar
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    Thanks, but could you explain how $\displaystyle \lim e^X=e^{\lim X}$ ?
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  8. #8
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    If $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle g$ is continuous on $\displaystyle L$, then $\displaystyle \lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a} f(x)\right) = g(L)$
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Greengoblin View Post
    Thanks, but could you explain how $\displaystyle \lim e^X=e^{\lim X}$ ?
    the rigorous theory might get a bit complicated, but it is somewhat similar to the fact that we can factor constants out side of limits. since the limit deals with X and not e, taking the limit of e to some power really only affects the power, since anything that is not a variable in X is not affected by the limit.

    EDIT: Chop Suey was a bit more rigorous .
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  10. #10
    Member Greengoblin's Avatar
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    Thanks, I got it now. I need to have a good look at limits again, as I haven't looked at them for ages and have forgotten alot.
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