# LImit with and exponent variable

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• February 26th 2011, 03:02 PM
jman919
LImit with and exponent variable
What is the limit of the sequence Attachment 20973as n approaches infinity.
• February 26th 2011, 03:15 PM
emakarov
Hint: $a_n=\left(1+\frac{2}{n+1}\right)^{n+1}/\left(1+\frac{2}{n+1}\right)$.
• February 26th 2011, 05:06 PM
jman919
Ok i understand how you got that.. but I'm not sure where to start. I keep getting that the limit is 1 which I know it is not.
• February 26th 2011, 05:19 PM
NOX Andrew
Here is a useful equation:

$\displaystyle \lim_{x\to\infty} \left ( 1 + \dfrac{r}{x} \right ) ^{x} = e^r$.

For your problem, instead of evaluating the limit of the ratio, evaluate the ratio of the limits:

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

The limit in the denominator is 1. Division by 1 won't change anything, so the answer to your original problem is just the limit in the numerator.

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

If we let r = 2 and x = n + 1, then the limit in your problem becomes $\displaystyle \lim_{x\to\infty} \left ( 1 + \dfrac{r}{x} \right ) ^{x}$. Remember the useful equation at the top?

The limit is just e^r. Substituting r = 2 gives: e^2.
• February 26th 2011, 06:34 PM
jman919
Thank you so much , it makes sense now.

My mistake was not evaluating the numerator correctly. I found the limit of the denominator correctly, but for the numerator I said that the limit when n approaches infinity was = (1)^ infinity which is the same as 1. What was wrong with this reasoning?
• February 26th 2011, 06:43 PM
NOX Andrew
$1^\infty$ is what is called an indeterminate form. It's value cannot be determined. In this problem, it turns out it is "equal" to e^2. In another problem, it might turn out to be "equal" to sin(1). Some other common indeterminate forms are $\frac{0}{0}$ and $\infty - \infty$. Here is a link to a Wolfram article on indeterminate forms: Indeterminate -- from Wolfram MathWorld
• February 26th 2011, 06:49 PM
jman919
Oh , thats right. I completely forgot that it was indeterminate.