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:

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

Here's how I solved it:

$\displaystyle \[\begin{array}{l}

{1^\infty }\,{\rm{is}}\,{\rm{indeterminate}}\\

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

Let\,f(x) = {e^x}\\

\[Let\,{a_n} = \ln {\left( {1 + \frac{1}{n}} \right)^n}\]

\end{array}\]$

$\displaystyle \[\begin{array}{l}

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

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

0 \cdot \infty \,{\rm{is}}\,{\rm{indeterminate}}

\end{array}\]$

$\displaystyle \[\begin{array}{l}

{\rm{Let }}t = \frac{1}{n}\\

{\rm{As}}\,n \to \infty ,t \to 0\\

\mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\ln \left( {1 + t} \right)}}{t}} \right]\\

\frac{0}{0}\,{\rm{is}}\,{\rm{indeterminate}}

\end{array}\]$

$\displaystyle \[\begin{array}{l}

\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\\

\mathop {\lim }\limits_{n \to \infty } f({a_n}) = f\left( {\mathop {\lim }\limits_{n \to \infty } {a_n}} \right) = f(1) = {e^1}\\

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

\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!