2
$\begingroup$

After doing a lot of homework, I've realized that I don't really get how to use the $\left(1+\frac{1}{n}\right)^n\to e$ method, which shows in how I can't solve indeterminate limits of the type $1^\infty$. For example:

$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{2}{n}\right)^n}$

Then I would do as follows:

$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{1}{2}+\frac{2}{n}-\frac{1}{2}\right)^n}$

$\lim_{n\to\infty}{\left(1+\frac{4-n}{2n}\right)^n}$

$\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}$

And, from here on, I would not know what to do. Thus, the question is:

How am I supposed to work with the exponents?

  • 2
    You do not need to use that limit. For n>8, note that ({1\over2}+{2\over n})^n< (3/4)^n. Use the Squeeze Theorem.2012-09-13

4 Answers 4

6

Write $\left(\frac{1}{2} + \frac{2}{n}\right)^n = \frac{1}{2^n} \left(1+\frac{4}{n}\right)^n$ Taking limits, we get $\lim_{n \rightarrow \infty} \left(\frac{1}{2} + \frac{2}{n} \right)^n = 0 \cdot e^4 = 0$

  • 0
    $\big(\frac{1}{2}+\frac{2}{n}\big)^n=\big(\frac{1}{2}(1+\frac{2/n}{1/2})\big)^n$.2012-09-13
1

Your final limit, $\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}\;,$ does not have the right form for you to apply the fact that $\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$, because the fraction $\frac{1}{\frac{2n}{4-n}}$ does not tend to $0$ as $n\to\infty$. Neither does the fraction $\dfrac{4-n}{2n}$ from the preceding step. Thus, either you need a different rearrangement altogether, or you need to use a different approach. As others have already suggested, a different approach does the job nicely.

1

Another approach:

If $\lim_{x\to{+\infty}} f(x)^{g(x)}$ is as $costant^{+\infty}$ which is indeterminate form then: $\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim_{x\to +\infty}\big(f(x)-1\big)g(x)}$

  • 0
    @amWhy: Hi! Amy. ;-)2013-03-18
0

Here's a trick:

$\lim_{n\rightarrow \infty}\left(\frac{1}{2}+\frac{2}{n}\right)^n =\lim_{n\rightarrow \infty}e^{n\ln\left(\frac{1}{2}+\frac{2}{n}\right)}$ which is $\,0\,$ since the limit of $n\ln\left(\frac{1}{2} + \frac{2}{n}\right)$ as n tends to infinity is $-\infty$.

Note, you can also use the trick above to prove that $\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n = e$