Solving for the Limit of a_p as p Goes to Infinity

  • Thread starter Thread starter mathjojo
  • Start date Start date
  • Tags Tags
    Infinity Limit
mathjojo
Messages
1
Reaction score
0
help! limit

find the limit as p goes to infinity

a_p = sqrt(p^2+p)-p

really don't know how to solve this... i know the limit is 1/2, but i need to prove that 1/2 is really the limit!1
 
Physics news on Phys.org
Have you realized the result of each of a_p=\sqrt{(p^2+p)}-p

Plug in the numbers 1 to 15? What do you get? What can you therefore tentatively assume about the limit? Now plug in the numbers from 15 to 30, what do you get? Again, what tentative assumption can you make? now 100 to 150?

As p increasesa_p becomes what?

What do you get with 1000? Now with 1001? Now with 1002?

Therefore what's the limit

\lim_{p\rightarrow\infty}

and why?
 
Last edited:
The numerical exercise can give moral justification to the answer.
However, to actually prove it, try the method of "conjugates". What is the conjugate of the function? Multiply and divide by the conjugate. The limit works out after some algebra.
 
I thought I had already replied to this:

One good way of getting rid of square roots is to multiply numerator and denominator by the "complement"- here \sqrt{p^2+ p}+ p.

Also, a good way to handle "p \leftarrow \infty is to divide numerator and denominator by a power of p so that you are left with 1/p which goes to 0.

(I can just imagine standing for your oral defense of your doctoral dissertation giving a "moral justification" of your result!)
 
Last edited by a moderator:
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