• Given$\lim_{x \to 0} \dfrac {a\sin hx + b \sin x}{x^3} $ is of the$\frac{0}{0}$ indeterminate form
• Thus applying L.Hospital rule,Differentiating numerator and denominator we have

$ \lim_{x \to 0} \dfrac {a\cos hx + b \cos x}{3x^2} \cdots\mathrm{Equation\ 1}$

• As x tends to 0 the denominator tends to 0 and numerator tends to a+b
• But as the limit is finite the numerator must also tend to 0
• Therefore a+b=0.......i
• Again from 1 we get by L Hospital rule

$\lim_{x \to 0} \dfrac {a\sin hx - b \sin x}{6x} $..$\frac{0}{0}$

• Applying L Hospital rule we have$\lim_{x \to 0} \dfrac {a\cos hx - b \cos x}{6} $
• Therefore$\dfrac{a-b}{6}=\dfrac{5}{3}$....ii
• Solving for a and b from i and ii we get

$a=5,b=-5 $