Ordered pair in limits

Both numerator and denominator tend to zero as $x \to 0$ ; so we can use L'Hopital's rule. Differentiating top and bottom, the limit is $\lim_{x \to 0} \frac{\sin 2x}{ae^{ax}-2b}$

Here, the numerator tends to zero, and the denominator tends to $a-2b$ . For the limit to be non-zero, we must have $a=2b$ ; so we can apply L'Hoptial again to get $\lim_{x \to 0} \frac{2\cos 2x}{a^2 e^{ax}}$

We can evaluate this at $x=0$ to get $\frac{2}{a^2}=\frac12$ , with solutions $a=\pm 2$ and $b=\pm 1$ (ie two solutions).