By the looks of it, I would say the following is a Neperian limit: $$\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}$$ but I could not find a way to algebraically bring it in the form: $$\lim_{x\to \infty}\biggl(1+\frac{k}{x}\biggr)^{mx} = e^{mk}$$
Any suggestion on how to solve this?
Hint: $$\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}=\left(\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{1/\sin(2/x^2)}\right)^{2\lim_{x\to \infty}\frac{\sin\frac2{x^2}}{\frac2{x^2}}}$$
Now for the exponent set $\dfrac2{x^2}=h\implies h\to0^+$