# Math Help - Help me find the limit? Is this an integral problem? I'm lost...

1. ## Help me find the limit? Is this an integral problem? I'm lost...

Compute the limit

limit as n approaches infinity of [1+(x/n)]^(n)

exactly, for each fixed x and use your calculator to write a decimal approximation of the limit when x = 1.

Basically, I don't know what the question is asking or how to approach the problem. If somebody could break it down for me that would be great.

We have a test tomorrow on integration and he wants us to use what we learned from that chapter.

Also, if somebody could provide a link that might benefit me I'd appreciate that too... the biggest problem I'm having is that I really don't know WHAT to google in this case. lol

Thanks guys.

2. Test it out.

$\left(1 + \frac{x}{1}\right)^{1}\;=\;1 + x$

$\left(1 + \frac{x}{2}\right)^{2}\;=\;1 + x + \frac{x^{2}}{4}$

$\left(1 + \frac{x}{4}\right)^{4}\;=\;1 + x + \frac{3x^{2}}{8}+ \frac{x^{3}}{16}+ \frac{x^{4}}{256}$

Are we getting anywhere? If we've done nothing else, it is hoped that you might be discouraged from determining the limit to be unuty, (1). It's more than 1+x, whatever it is.

We could also try the Binomial Theorem on it. Is that in your bag of tricks?

OR, you could find the limit of the logarithm and see where that leads.

$log[(1+x/n)^{n}] = n\cdot log(1+x/n) = \frac{log(1+x/n)}{\frac{1}{n}}$

Your task is to determine why on earth I would write it in that last form.

3. I would imagine that he wants us to use the last method you provided.

And the reason you would do so would be to relocate the nth power to the front of the equation, right?

4. Also, using the first method... I don't know the binomial theorem well enough to use it, but taking the limit of the equation to the nth power... through using l'Hopitals Rule, would the solution be x/2?

The last bit of the equation would be limit (x^(n))/(n^(2)) =LH= limit nx/2n = x/2?

5. It seems like you are on the edge of understanding. The purpose of the last form is to demonstrate the applicability of l'Hopital.

Rewrite the argument of the logarithm, and then try l'Hopital.

$
\frac{log(\frac{n+x}{n})}{\frac{1}{n}}
$