Finding xn as n Tends to Infinity: Q1 & Q2

  • Thread starter Thread starter Mattofix
  • Start date Start date
  • Tags Tags
    Infinity
Mattofix
Messages
137
Reaction score
0

Homework Statement



2 Questions, both find xn as n tends to infinity.

http://img229.imageshack.us/img229/5154/scan0002un5.th.jpg

Homework Equations


The Attempt at a Solution



Have attempted question one but am unsure if (1/n)log(n^2) tends to 0, and if it does do i need to prove it? I don't know how to do the second q, i know that sin(expn) oscillates between -1 and 1 and exp(-n) tends to 0 as n tends to infinity
 
Last edited by a moderator:
Physics news on Phys.org
Yes, (1/n) log(n^2) = (2/n)log(n) goes to 0. You might prove that by looking at 2ln(x)/x^2 and using L'Hopital's rule.

As for the second one, since sin is always between -1 and 1, you really just need to show that \sqrt{n}/(n+ e^{-n})< \sqrt{n}/n (since e^{-n} is always positive) converges to 0.
 
thanks :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top