As Dr. Peterson said, since the denominator goes to 0, in order that the limit exist, the numerator must also go to 0 and that makes it a candidate for "L'Hopital's rule". The derivative of the numerator, with respect to x, is \(\displaystyle -b sin(x)+ 4ae^{4x}- ce^{-x}\) and the derivative of the denominator, with respect to x, is -sin(2x)-2x cos(x). The denominator goes to 0 so, in order that the limit exist, the numerator must also. That gives a second equation for a, b, and c and allows us to use "L'Hopital" a second time. Differentiating the numerator again, we have \(\displaystyle -b cos(x)+ 16ae^{4x}+ ce^{-x}\). Differentiating the denominator again, \(\displaystyle -4 cos(3x)+ 2x sin(x)\).
Finally, as x goes to 0, the denominator goes to -4, not 0 and the numerator goes to -b+ 16a+ c so we must have \(\displaystyle -\frac{-b+ 16a+ c}{4}= 10\).
We have three equations to solve for a, b, and c.