Applying L'Hospital's Rule,

$$ \therefore \lim_{x\to 0} \dfrac{asinh x + b sinx}{x^3} \;=\; \lim_{x\to 0} \dfrac{acosh x + b cosx}{3x^2} $$

Now here,as x→0,denominator tends to zero.

Hence,if the limit is to be finite,we must have numerator tending to zero.

$\therefore a cosh⁡(0)+b cos⁡(0)=0, \; i.e. \; a+b=0 \; \; \ldots(i) $

Now,again applying L'Hospital' s Rule,

$$ \therefore \lim_{x\to 0} \dfrac{asinh x - b sinx}{6x} \;=\; \dfrac{5}{3} $$

$$ \therefore \lim_{x\to 0} \dfrac{acosh x - b cosx}{6} \;=\; \dfrac{5}{3} $$

Substituting x=0,

$ \therefore a cosh⁡(0)-b cos⁡(0) \;= \; \dfrac{6 \times 5}{3} \; \; , \; \; i.e. \; a-b=10 \; \; \; … \; (ii) $

Solving simultaneously equations (i)and (ii),we get a=5 and b=-5