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lemon14 · Nov 24, 2011, 08:41 AM

Could you give me some advice on how to approach this type of problems, please?

$a,b \epsilon R$

$1. \lim_{x \to \infty} \left(a \frac{bn}{n^2-1} \right)^n = e^2$

$2. \lim_{x \to \infty} \left(\frac{n^2 an 1}{n^2 3n -2} \right)^n = e$

$3. \lim_{x \to \infty} \left(\frac{an^2 bn 2}{bn^2 4n 3} \right)^{\frac{n^2}{n 1}} = \frac{1}{e}$

$4. \lim_{x \to \infty} (\sqrt{n^2 n 1} - \sqrt{n^2 an 2})^{\frac{bn^2 n}{n 1}} = \frac{1}{e}$

$5. \lim_{x \to \infty} (\sqrt[n]{a} - \sqrt[2n]{a}) = 1, a>0$

lemon14 · Nov 24, 2011, 08:55 AM

Sorry, there are some mistakes. It seems there occurred an error with "+".

1. lim [a + bn/(n^2 - 1)]^n = e^2

2. lim [(n^2 + an + 1)/(n^2 + 3n - 2)]^n = e

3. lim [(an^2 + bn + 2)/(bn^2 + 4n + 3)]^[n^2/(n+1)] = 1/e

4. lim [sqrt(n^2 + n + 1) - sqrt(n^2 + an + 2)]^[(bn^2 + n)/(n+1)] = 1/e

lemon14 · Nov 24, 2011, 08:59 AM

... and $\lim_{n \to \infty}$

corrigan · Nov 26, 2011, 09:56 PM

lemon14,

a,b[\epsilon R?] just means a and b are real numbers. Wikipedia has a pretty good exposition on set theoretic notation http://en.wikipedia.org/wiki/Naive_set_theory

As for the rest, I'm assuming you are solving for a and b. I won't do your homework for you, but I did number 2 to give you an idea on the rest. Let me know if it helps. :)

lemon14 · Nov 27, 2011, 01:17 AM

Thank you Corrigan. It is helpful, indeed, although I haven't learned about L'Hopital's rule yet.

I am getting started with the rest of the exercises, but I'm not sure about the those with both a and b. Should I set a value for one of them in order to find the other one? I'm not asking you to solve me another exercise, for I know it takes you pretty much time, I just need a starting point.

corrigan · Nov 27, 2011, 08:00 AM

L'hopital's rule says if lim u = lim v = 0 or infinity and lim (u / v) exists, then lim (u / v) =lim (du / dv). Where du and dv are the derivatives.

I haven't worked out the others, but when you have two variables, you'll probably get a in terms of b, from there it's just algebra. With all of these problems the thing to keep in mind is that $\lim_{n \to \infity} (1 + \frac{1}{n})^n = e$ .

corrigan · Nov 27, 2011, 08:02 AM

wow that didn't display at all like I wanted it to. Okay, e is defined to be the limit as n approaches infinity of (1 + (1/n))^n.

lemon14 · Nov 27, 2011, 01:58 PM

wow that didn't display at all like i wanted it to. Okay, e is defined to be the limit as n approaches infinity of (1 + (1/n))^n.

For a mathematical display you should write math mathematical expression /math. "math" words must be between []

corrigan · Nov 27, 2011, 03:13 PM

/math[\lim_{n \to \infity} (1 + \frac{1}{n})^n = e] ?

corrigan · Nov 27, 2011, 03:19 PM

[/lim]?

corrigan · Nov 27, 2011, 03:22 PM

$\lim_{n \to \infity} (1 + \frac{1}{n})^n = e$ ?

corrigan · Nov 27, 2011, 03:22 PM

Got it.