Differential Calc (Math)

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Stuck... can't seem to use l'hopitals, nor can I use logrithmic differentiation...

lim (e^(2x)+ x)^(1/3x)

Thank you so much

  • Differential Calc (Math) -

    If that's ^(1/3)x, then you have no problem:

    (e^0+0)^(0) = (1+0)^0 = 1^0 = 1

    Now, if it's ^(1/(3x)) then you have
    1^∞, so we have to get creative.

    How about:

    e^2x = 1 + 1/1(2x) + 1/2(2x)^2 + ...
    = 1+2x as x->0 because the higher powers become insignificant.

    lim (1+3x)^1/(3x) = lim (1+u)^1/u = e

    Can't think of something more rigorous at the moment.

  • Differential Calc (Math) -

    I'm not quiet sure how e^2x equals what you put,


    lim (1+3x)^1/(3x) = lim (1+u)^1/u = e

    doesn't seem to make sense either..

    Maybe a more dumbed down version? And it is in fact ^1/(3x)

  • Differential Calc (Math) -

    the Taylor series for e^u = 1 + 1/1! u + 1/2! u^2 + 1/3! u^3 + ...

    the definition of e is lim(u->0) (1+u)^1/u

  • Differential Calc (Math) -

    Hm... Never learned this before, so I'm not quite sure if I can use this.

    However, I appreciate the effort!

  • Differential Calc (Math) -

    How about this:

    lim(x->0) u^v = lim(x->0) e^(v ln u)

    here we have
    u = e^2x + x
    v = 1/(3x)

    lim e^(ln(e^2x+x)/3x)
    = e^ [lim(ln(e^2x+x)/3x)]
    now use l'Hospital's Rule to see that

    lim ln(e^2x+x)/3x = (2e^2x+1)/(e^2x+x) / 3
    = (2+1)/(0+1) / 3
    = 3/3 = 1

    and our limit is now

    e^(1) = e

  • Differential Calc (Math) -

    WOW! Thank you so much! You rock!

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