Limit of a Bessel function is a Gaussian

Question

Let $x$ and $\nu$ be real. Is the following true?

$$ e^{-\frac{x^2}{4}}=\lim_{\nu\rightarrow\infty}\Gamma(\nu+1)\left(\frac{2}{\sqrt\nu x}\right)^\nu J_\nu\left(\sqrt{\nu} x\right), $$

It appears to be true numerically, but I couldn't prove it.

Answer 1

Use the first term of Gradshteyn & Rhyzhik 8.452 'approximation by tangents:' $$ J_v\Big(\frac{v}{\cosh(a)}\Big) \sim \frac{\exp{(v \tanh(a) - v\,a)}} {\sqrt{2\, v\,\pi \tanh(a) }} $$ Set $\cosh(a) = \sqrt{v}/x $ for fixed $x,$ so that this parameter is increasing, a necessity to use the formula. Then $\tanh{a} = \sqrt{1-\text{sech }^2{ a}} = \sqrt{1-x^2/v} \to 1.$ Use $a=\text{arccosh(}\cosh(a)) \sim \log{\big(2\sqrt{v}/x \big)}-x^2/v.$ Use $\Gamma(v+1) \sim v^{v}\sqrt{2\pi\,v} \, e^{-v} .$ Substitute this into the right hand side of the equation, and algebra completes the proof.