Trivial limit ( 1 - (-x)^n ) / 1 + x

  • Thread starter Thread starter davitz
  • Start date Start date
  • Tags Tags
    Limit
davitz
Messages
2
Reaction score
0

Homework Statement



lim ( 1 - ( - x ) ^ n ) / ( 1 + x ) as n -> infinity

Homework Equations



I can't understand why this equals to 1 / ( 1 + x ) (No matter what power of " n " was x e.q: x ^ 2n or x ^ ( n ^ 2 )

The Attempt at a Solution



I have no clue what rule to apply. I thought it might be a case of using the lim ( 1 + 1 / n ) ^ n to get to " e " but this seems like a non-sense in this case.
 

Attachments

  • equation.png
    equation.png
    738 bytes · Views: 517
Physics news on Phys.org
I think you first need to have a condition on x before that is true.

0 < x < 1, right?

If so, anything between 0 and 1 raised to a power of infinity will tend to 0, as it gets smaller with each successive multiplication.

\lim_{n\to \infty} x^n = 0

where -1<x<1
 
Thank so much for the reply! Yes, x > -1 and x < 1 or -1 < x < 1 and now I understand why this is the result!

Thank you again!
 
Glad to have been of help! :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