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 Jan16 comment Is odd continuous function differentiable at $x=0$? N.S., you are right! $\frac{f(x)}{x} = \sin\frac{1}{x^2}$ which oscillates between -1 and 1 and so the limit does not exist. So this function is not differentiable at $x=0$. Thanks! Jan16 asked Is odd continuous function differentiable at $x=0$? Oct25 awarded Tumbleweed Oct10 asked How to prove Gegenbauer's addition theorem? Sep21 awarded Scholar Sep21 accepted How to prove asymptotic limit of an incomplete Gamma function Sep21 comment How to prove asymptotic limit of an incomplete Gamma function I verified your steps, I think it's all correct. I am accepting your answer as it gives a simple proof. Sep21 comment How to prove asymptotic limit of an incomplete Gamma function Thanks! As a matter of fact, I actually wanted to prove that $\gamma(z, x)\over\Gamma(z)$ goes to zero, but thought it'd be easier to do it with the upper incomplete gamma function. Sep21 comment How to prove asymptotic limit of an incomplete Gamma function So you proved that $\gamma(z, x) < {x\over z-1-x}\Gamma(z)$ for $z-1 > x > 0$ and from that the limit follows immediately. Nice! Sep21 asked How to prove asymptotic limit of an incomplete Gamma function May9 awarded Student May9 comment How to prove that Legendre polynomials form a complete basis using functional analysis Ah -- I forgot about the boundary condition -- so requiring the solutions to satisfy some conditions (for example being non-singular, or I can even require something stronger, like being $\pm 1$ at the endpoints), then the only allowed $\lambda$ are of the form $n(n+1)$, thus proving that the spectrum is discrete. If we allow Legendre functions, then the spectrum is continuous --- because I think that the Legendre functions can be normalized to (Dirac) delta function (using physics terminology). What about 2? May9 awarded Supporter May9 comment How to prove that Legendre polynomials form a complete basis using functional analysis @J.M., for non-integer $n$ (i.e. a general $\lambda$), the solutions are Legendre functions (en.wikipedia.org/wiki/Legendre_function). So it would seem that all $\lambda$ are actually allowed, thus the spectrum being continuous... May9 asked How to prove that Legendre polynomials form a complete basis using functional analysis