If $$\beta = \displaystyle\lim_{x \to 0} \dfrac{e^{x^3} - (1-x^3)^{1/3} + \left((1-x^2)^{1/2} - 1\right)\sin x}{x \sin^2 x}$$, then the value of $$6\beta$$ is _______.

Let

$$N(x)=e^{x^{3}}-\left(1-x^{3}\right)^{1/3}+\bigl((1-x^{2})^{1/2}-1\bigr)\sin x,$$
$$D(x)=x\sin^{2}x.$$

We need $$\beta=\displaystyle\lim_{x\to 0}\dfrac{N(x)}{D(x)}.$$

Step 1 : Expand every term about $$x=0$$

Maclaurin series that will be used:

• $$e^{t}=1+t+\dfrac{t^{2}}{2}+\dots$$
• $$(1+u)^{n}=1+nu+\dfrac{n(n-1)}{2}u^{2}+\dots$$
• $$\sin x = x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120}+\dots$$

(i) $$e^{x^{3}} = 1+x^{3}+\dfrac{x^{6}}{2}+O(x^{9}).$$

(ii) $$(1-x^{3})^{1/3}=1+\dfrac{1}{3}(-x^{3})+\dfrac{1/3(-2/3)}{2}(-x^{3})^{2}+O(x^{9}) =1-\dfrac{x^{3}}{3}-\dfrac{x^{6}}{9}+O(x^{9}).$$

Hence

$$e^{x^{3}}-\left(1-x^{3}\right)^{1/3} =(1+x^{3}+\tfrac{x^{6}}{2})-(1-\tfrac{x^{3}}{3}-\tfrac{x^{6}}{9})+O(x^{9}) =\dfrac{4}{3}x^{3}+\dfrac{11}{18}x^{6}+O(x^{9}).$$

(iii) $$(1-x^{2})^{1/2}=1-\dfrac{x^{2}}{2}-\dfrac{x^{4}}{8}+O(x^{6}),$$
so $$(1-x^{2})^{1/2}-1=-\dfrac{x^{2}}{2}-\dfrac{x^{4}}{8}+O(x^{6}).$$

Multiply by $$\sin x$$:

$$\bigl((1-x^{2})^{1/2}-1\bigr)\sin x =\Bigl(-\dfrac{x^{2}}{2}-\dfrac{x^{4}}{8}+O(x^{6})\Bigr) \Bigl(x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120}+O(x^{7})\Bigr).$$

Keep terms up to $$x^{5}$$ (higher-order terms will vanish upon taking the limit):

$$$ \begin{aligned} &-\dfrac{x^{2}}{2}\cdot x =-\dfrac{x^{3}}{2},\\ &-\dfrac{x^{2}}{2}\cdot\Bigl(-\dfrac{x^{3}}{6}\Bigr)=+\dfrac{x^{5}}{12},\\ &-\dfrac{x^{4}}{8}\cdot x =-\dfrac{x^{5}}{8}. \end{aligned} $$$

Adding these,

$$\bigl((1-x^{2})^{1/2}-1\bigr)\sin x =-\dfrac{x^{3}}{2}-\dfrac{x^{5}}{24}+O(x^{7}).$$

(iv) Combine parts (i)-(iii):

$$$ \begin{aligned} N(x)&=\Bigl(\dfrac{4}{3}x^{3}+\dfrac{11}{18}x^{6}\Bigr) +\Bigl(-\dfrac{x^{3}}{2}-\dfrac{x^{5}}{24}\Bigr)+O(x^{7})\\ &=\dfrac{5}{6}x^{3}-\dfrac{x^{5}}{24}+\dfrac{11}{18}x^{6}+O(x^{7}). \end{aligned} $$$

(v) Now expand the denominator:

$$\sin x = x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120}+O(x^{7}),$$

so

$$\sin^{2}x =\bigl(x-\dfrac{x^{3}}{6}+\dfrac{x^{5}}{120}\bigr)^{2} =x^{2}-\dfrac{x^{4}}{3}+\dfrac{2x^{6}}{45}+O(x^{8}).$$

Therefore

$$D(x)=x\sin^{2}x =x^{3}-\dfrac{x^{5}}{3}+\dfrac{2x^{7}}{45}+O(x^{9}).$$

Step 2 : Form the quotient

Write both series factored by $$x^{3}$$:

$$N(x)=x^{3}\Bigl(\dfrac{5}{6}-\dfrac{x^{2}}{24}+\dfrac{11x^{3}}{18}+O(x^{4})\Bigr),$$

$$D(x)=x^{3}\Bigl(1-\dfrac{x^{2}}{3}+\dfrac{2x^{4}}{45}+O(x^{6})\Bigr).$$

Hence

$$\dfrac{N(x)}{D(x)} =\dfrac{\dfrac{5}{6}-\dfrac{x^{2}}{24}+O(x^{3})} {1-\dfrac{x^{2}}{3}+O(x^{4})}.$$

Using $$\dfrac{1}{1-z}=1+z+O(z^{2})$$ for small $$z$$,

$$$ \begin{aligned} \frac{N(x)}{D(x)} &=\Bigl(\dfrac{5}{6}-\dfrac{x^{2}}{24}+O(x^{3})\Bigr) \Bigl(1+\dfrac{x^{2}}{3}+O(x^{4})\Bigr)\\[4pt] &=\dfrac{5}{6}+O(x^{2}). \end{aligned} $$$

Step 3 : Take the limit

As $$x\to 0$$, the higher-order terms vanish, giving

$$\beta=\dfrac{5}{6}.$$

Therefore

$$6\beta = 6\left(\dfrac{5}{6}\right)=5.$$

Final Answer: 5