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Basic Limit Question

  1. Dec 19, 2009 #1
    This is a pretty basic limit question regarding the limit,

    [tex] \lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = e [/tex]

    Wolframalpha gives the following reasoning for this answer:

    [tex] \lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = e^{\lim_{x \rightarrow \infty} x\ln{(1+\frac{1}{x})}} = e^{\lim_{t \rightarrow 0} \frac{\ln{(1+t)}}{t}} = e^{\lim_{t \rightarrow 0} \frac{ \frac{d\ln{(1+t)}}{dt}}{\frac{d}{dt}t}}= e^{\lim_{t \rightarrow 0} \frac{1}{1+t}} = e^1 [/tex]

    My question is, by the same reasoning, why is the following not true? (where log is log base 10)

    [tex] \lim_{x \rightarrow \infty} (1+\frac{1}{x})^x = 10^{\lim_{x \rightarrow \infty} x\log{(1+\frac{1}{x})}} = 10^{\lim_{t \rightarrow 0} \frac{\log{(1+t)}}{t}} = 10^{\lim_{t \rightarrow 0} \frac{ \frac{d\log{(1+t)}}{dt}}{\frac{d}{dt}t}}= 10^{\lim_{t \rightarrow 0} \frac{1}{1+t}} = 10^1 [/tex]

    Am I missing something??
     
  2. jcsd
  3. Dec 19, 2009 #2
    Wooooow, hold on, I understand what I did wrong...

    The derivative of [tex] \log{(1+t)} [/tex] is NOT [tex] \frac{1}{1+t} [/tex] but rather [tex] \frac{1}{(1+t)\ln{10}} [/tex]

    Sorry for wasting the time of those who've read this.
     
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