Vtyjkyuk

Prove that: $$limlimits_mto infty fracintlimits_0^pi/2(sin(x))^2m dxintlimits_0^pi/2(sin(x))^2m+1 dx = 1$$

I had tried to get it from inequalities like $$(sin(x))^2m+1 le (sin(x))^2m le (sin(x))^2m-1$$
But I always hit a wall in that regard (I integrate the sin functions and from there keep on going but there it ends).

You can calculate the integrals easily by using the first reduction formula at http://www.vias.org/calculus/07_trigonometric_functions_05_03.html. Notice that the term outside the integral becomes $0$ when the limits are integration are substituted.

You were on the right track. Let $I_n = int_0^pi/2sin(x)^n,dx$. Then $I_n_ngeq 0$ is clearly a decreasing sequence convergent to zero, and by integration by parts
$$ I_2m+2 = frac2m+12m+2,I_2m$$
In particular
$$ fracI_2mI_2m+1 geq fracI_2m+1I_2m+1 = 1 $$
and
$$ fracI_2mI_2m+1 leq fracI_2mI_2m+2 = 1+frac12m+1, $$
so the claim follows by squeezing.
You may also prove $fracI_2mI_2m+1sim 1+frac28m+3$ through essentially the same technique.

Just perform the integrals.

The numerator is:

$$fracsqrtpi Gamma left(m+frac12right)2 Gamma (m+1)$$

and the denominator is

$$fracsqrtpi Gamma (m+1)2 Gamma left(m+frac32right)$$

so the limit arises directly.

If you have a look at Table of Integrals, Series, and Products (Seventh Edition) by I.S. Gradshteyn and I.M. Ryzhik, you should find that
$$intlimits_0^pi/2(sin(x))^2m, dx=frac(2m-1)!! (2m)!! frac pi 2=fracsqrtpi , Gamma left(m+frac12right)2, Gamma (m+1)$$
$$intlimits_0^pi/2(sin(x))^2m+1, dx=frac(2m)!! (2m+1)!! =fracsqrtpi , Gamma (m+1)2 ,Gamma left(m+frac32right)$$ (formulae FI II 151). This makes
$$I_m=fracintlimits_0^pi/2(sin(x))^2m, dxintlimits_0^pi/2(sin(x))^2m+1, dx= fracGamma left(m+frac12right) Gamma left(m+frac32right)Gamma
(m+1)^2$$ Now, take logarithms, use Stirling approximation and continue with Taylor series for large values of $m$ to get
$$log(I_m)=frac14 m-frac18 m^2+Oleft(frac1m^3right)$$
$$I_m=e^log(I_m)=1+frac14 m-frac332 m^2+Oleft(frac1m^3right)$$ which, for sure, shows the limit but also how it is approached.

As shown in the table below, the approximation is quite good even for small values of $m$.
$$left(
beginarrayccc
m & textexact & textapproximation \
1 & 1.17810 & 1.15625 \
2 & 1.10447 & 1.10156 \
3 & 1.07379 & 1.07292 \
4 & 1.05701 & 1.05664 \
5 & 1.04644 & 1.04625 \
6 & 1.03917 & 1.03906 \
7 & 1.03387 & 1.03380 \
8 & 1.02983 & 1.02979 \
9 & 1.02665 & 1.02662 \
10 & 1.02409 & 1.02406
endarray
right)$$

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beginalign
int_0^pi/2sin^nparsx,dd x & =
int_0^pi/2cos^nparsx,dd x =
int_0^pi/2expparsnlnparscosparsx,dd x
\[5mm] & stackrelmrmas n to inftysim
int_0^inftyexpo-nx^2/2
pars1 - x^4 over 12,dd x
quadpars~Laplace Method~
\[5mm] & = rootpi over 21 over rootn -
1 over 4rootpi over 21 over n^5/2
endalign