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Math Help - Is there another way to do this sequence limit problem?

  1. #1
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    Is there another way to do this sequence limit problem?

    Hi again all,

    I'm just looking at a sequence limit problem, and I'm wondering if there's a different way to do it than I did. Here it is:

    \[\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n}\]

    Here's how I solved it:

    \[\begin{array}{l}<br />
{1^\infty }\,{\rm{is}}\,{\rm{indeterminate}}\\<br />
\mathop {\lim }\limits_{n \to \infty } {e^{\ln {{\left( {1 + \frac{1}{n}} \right)}^n}}}\\<br />
Let\,f(x) = {e^x}\\<br />
\[Let\,{a_n} = \ln {\left( {1 + \frac{1}{n}} \right)^n}\]<br />
\end{array}\]
    \[\begin{array}{l}<br />
\mathop {\lim }\limits_{n \to \infty } \left[ {\ln {{\left( {1 + \frac{1}{n}} \right)}^n}} \right]\\<br />
\mathop {\lim }\limits_{n \to \infty } \left[ {n\ln \left( {1 + \frac{1}{n}} \right)} \right]\\<br />
0 \cdot \infty \,{\rm{is}}\,{\rm{indeterminate}}<br />
\end{array}\]
    \[\begin{array}{l}<br />
{\rm{Let }}t = \frac{1}{n}\\<br />
{\rm{As}}\,n \to \infty ,t \to 0\\<br />
\mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\ln \left( {1 + t} \right)}}{t}} \right]\\<br />
\frac{0}{0}\,{\rm{is}}\,{\rm{indeterminate}}<br />
\end{array}\]
    \[\begin{array}{l}<br />
\mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\ln \left( {1 + t} \right)}}{t}} \right]\mathop  = \limits^{L'H} \mathop {\lim }\limits_{t \to 0} \left[ {\frac{1}{{1 + t}}} \right] = 1\\<br />
\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right) = f(1) = {e^1}\\<br />
\[\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n} = e\]<br />
\end{array}\]

    Would you say this was the best approach, or was there an easier way to go about this? The reason why I ask is I had to scratch my head for awhile before I realized I could address the limit near the end with a substitution.

    Thanks in advance!
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  2. #2
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    It is well known that \displaystyle \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e, but if you didn't know this...

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

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

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

    \displaystyle = e^{\lim_{n \to \infty}\frac{\ln{\left(1 + \frac{1}{n}\right)}}{\frac{1}{n}}}.


    Now, this is a \displaystyle \frac{0}{0} indeterminate form, so use L'Hospital's Rule...
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  3. #3
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    I'm still very new to all of this (if that isn't obvious), and it's been a learning curve to say the least with self-study, but that makes sense. It's pretty much the same thing that I did except without the substitution. (just putting it in terms of 1/n instead of replacing it).

    Since it's so well known, what's the usual application you see with it?
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  4. #4
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    Compound interest.
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  5. #5
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    Ah I see. Well, thanks!

    I must say that sequences and series are absolutely fascinating.
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