Does (a_n)^{b_n} converge to a^b as n approaches infinity?

  • Thread starter Thread starter no_alone
  • Start date Start date
no_alone
Messages
32
Reaction score
0

Homework Statement


an -> a
bn -> b
prove that anbn = ab
?
Its all Sequence of course
 
Physics news on Phys.org
Hint: (a_n)^{b_n} = e^{b_n log(a_n)}

Now use continuity of the exponential and logarithm function to take the limits "inside the function".
 
By the way, we don't want to prove that (a_n)^{b_n} = a^b. We want to prove that (a_n)^{b_n} \rightarrow a^b as n \rightarrow \infty
 
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