What is the limit for following expression
$$f(x)=\lim_{m\to 0}\frac{{}_1F_2\left(-\dfrac 12;\dfrac 12,\dfrac m2;x\right)}{\Gamma(m)}\quad \text{with}\quad x\in\mathbb{R},$$
where ${}_1F_2$ is the hypergeometric series and $\Gamma$ the gamma function? Single values are
$$f(-1) \approx 1.69809, \qquad f(0)=0, \qquad f(1) \approx-2.36873$$
First we write out the function with the Pochhammer symbol:
$$\lim_{m\to0}\frac{\,_1\text F_2\left(-\frac12;\frac12,\frac m2,x\right)}{\Gamma(m)}=\lim_{m\to0}\sum_{n=0}^\infty \frac{\left(-\frac12\right)_nx^n}{\left(\frac12\right)_n\left(\frac m2\right)_n\Gamma(m)n!}$$
A limit gives:
$$f(x)=\lim_{m\to0}\sum_{n=0}^\infty \frac{\left(-\frac12\right)_nx^n}{\left(\frac12\right)_n\left(\frac m2\right)_n\Gamma(m)n!}=\sum_{n=0}^\infty\frac{2\left(-\frac12\right)_nx^n}{n!\Gamma(n)\left(\frac12\right)_n}=-2x\,_1\text F_2\left(\frac12;\frac32,2;x\right) $$
It is a combination of Struve L and Bessel I functions:
$$\lim_{m\to0}\frac{\,_1\text F_2\left(-\frac12;\frac12,\frac m2,x\right)}{\Gamma(m)}=2\text I_1\left(2\sqrt x\right)\left(\pi x\text L_0\left(2\sqrt x\right)+\sqrt x\right)-2x\text I_0 \left(2\sqrt x\right) \left(\pi \text L_1 \left(2\sqrt x\right)+2\right)$$
When evaluating, Bessel J and Struve H appear because of the imaginary unit. Single values include
$$f(-1)=2\text J_1(2)(\pi\text H_0(2)-1)-2\text J_0(2)(\pi \text H_1(2)-2)= 1.69809097088030636354471441251561 \\f(1)=2\text I_1(2)(\pi\text L_0(2)+1)-2\text I_0(2)(\pi\text L_1(2)+2)= -2.36873010158184893306470337240… $$