# Finding the Limit

• April 11th 2012, 09:16 AM
spruancejr
Finding the Limit
Given lim $S_n=(1+\dfrac{1}{n})^n \rightarrow e$

Find
1)lim $S_n=(1+\dfrac{1}{2n})^{2n} \rightarrow ?$

2)lim $S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?$

Attempt:
I see that $\dfrac{1}{2n}$ converges to zero which makes it no different than $\dfrac{1}{n}$ so we're left with $S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?$ which should make both problems #1 and #2 have a limit of $e^2$.

However, the back of the book indicates that #1 has a limit of e and #2 has a limit of $e^2$

Help?
• April 11th 2012, 09:36 AM
Plato
Re: Finding the Limit
Quote:

Originally Posted by spruancejr
Given lim $S_n=(1+\dfrac{1}{n})^n \rightarrow e$
Find
1)lim $S_n=(1+\dfrac{1}{2n})^{2n} \rightarrow ?$

2)lim $S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?$

Attempt:
I see that $\dfrac{1}{2n}$ converges to zero which makes it no different than $\dfrac{1}{n}$ so we're left with $S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?$ which should make both problems #1 and #2 have a limit of $e^2$.

However, the back of the book indicates that #1 has a limit of e and #2 has a limit of $e^2$

The text is correct in both cases.
Here is a simple rule: ${\left( {1 + \frac{a}{{bn + c}}} \right)^{dn}} \to {e^{\frac{{ad}}{b}}}$
• April 11th 2012, 10:07 AM
spruancejr
Re: Finding the Limit
Thanks!
• April 11th 2012, 05:57 PM
Prove It
Re: Finding the Limit
Quote:

Originally Posted by spruancejr
Given lim $S_n=(1+\dfrac{1}{n})^n \rightarrow e$

Find
1)lim $S_n=(1+\dfrac{1}{2n})^{2n} \rightarrow ?$

2)lim $S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?$

Attempt:
I see that $\dfrac{1}{2n}$ converges to zero which makes it no different than $\dfrac{1}{n}$ so we're left with $S_n=(1+\dfrac{1}{n})^{2n} \rightarrow ?$ which should make both problems #1 and #2 have a limit of $e^2$.

However, the back of the book indicates that #1 has a limit of e and #2 has a limit of $e^2$

Help?

Your book is correct. For 1) if you let \displaystyle \begin{align*} t = 2n \end{align*} and note that as \displaystyle \begin{align*} n \to \infty, t \to \infty \end{align*}, then \displaystyle \begin{align*} \lim_{n \to \infty}\left(1 + \frac{1}{2n}\right)^{2n} = \lim_{t \to \infty}\left(1 + \frac{1}{t}\right)^t \end{align*}, which is the exact same limit as you were given in the first place.

The reason 2) is \displaystyle \begin{align*} e^2 \end{align*} is because \displaystyle \begin{align*} \left(1 + \frac{1}{n}\right)^{2n} = \left[\left(1 + \frac{1}{n}\right)^n\right]^2 \end{align*}.