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Limit with e

  1. Jan 14, 2009 #1
    how do i solve this limit

    Lim (ex+sinx)1/sinx
    x->0

    its almost oiler (1+x)1/x, at 1st looking at it i thought it was simple e, because ex when x->0 is 1, but i see i need to somehow turn the ex into 1,,, any ideas??
     
  2. jcsd
  3. Jan 14, 2009 #2
    I would take ln of both sides and then use L'Hopital's
     
  4. Jan 14, 2009 #3
    both of which sides?? can you show/describe what you mean
     
  5. Jan 14, 2009 #4
    take the ln and use L'Hopital.

    If you just want to evaluate it, expand it first order in x should be good.

    Be careful though, for small x,
    [tex]e^x\approx 1+x[/tex]
    and
    [tex]\sin(x) \approx x[/tex]
    so your "intuitive" conclusion (1+x)^{1/x} doesn't hold.
     
  6. Jan 14, 2009 #5
    what do i do with the ln,?? take ln on what??
     
  7. Jan 14, 2009 #6
    You have

    [tex] y = \lim_{x \rightarrow 0} \left(e^{x} + sin(x)\right)^{\frac{1}{sin(x)}} [/tex]

    If you take ln of both sides of the expression you have

    [tex] ln(y) = \lim_{x \rightarrow 0} \frac{ln(e^{x} + sin(x))}{sin(x)} [/tex]

    Now do L'Hopital's and go from there, remember in the end you want to solve for y.
     
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