# Thread: Is there another way to do this sequence limit problem?

1. ## 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:

$\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.

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

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

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

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

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

Now, this is a $\displaystyle \displaystyle \frac{0}{0}$ indeterminate form, so use L'Hospital's Rule...

3. 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?

4. Compound interest.

5. Ah I see. Well, thanks!

I must say that sequences and series are absolutely fascinating.