Question

If $\beta=\displaystyle\lim _{x \rightarrow 0} \frac{e^{x^3}-\left(1-x^3\right)^{\frac{1}{3}}+\left(\left(1-x^2\right)^{\frac{1}{2}}-1\right) \sin x}{x \sin ^2 x}$ then the value of $6 \beta$ is ____.

Answer 1

The answer is 5.

Answer 2

Given:
\(\beta = \lim_{x \to 0} \frac{e^{x^3} - (1 - x^3)^{\frac{1}{3}} + \left( (1 - x^2)^{\frac{1}{2}} - 1 \right) \sin x}{x \sin^2 x}\)

1. Examine \(e^{x^3}\):
For \(x \to 0\),
\(e^{x^3} \approx 1 + x^3 \quad \text{(ignoring higher-order terms)}\)

2. Examine \((1 - x^3)^{\frac{1}{3}}\):
Using the binomial approximation for small x,
\((1 - x^3)^{\frac{1}{3}} \approx 1 - \frac{x^3}{3}\)

3. Examine \((1 - x^2)^{\frac{1}{2}}\):
Using the binomial approximation,
\((1 - x^2)^{\frac{1}{2}} \approx 1 - \frac{x^2}{2}\)  So,

\((1 - x^2)^{\frac{1}{2}} - 1 \approx - \frac{x^2}{2}\)

4. Examine sin x:
For small x,
\(\sin x \approx x\)

Now, substitute these approximations back into the original limit expression:

\(e^{x^3} \approx 1 + x^3\)
\((1 - x^3)^{\frac{1}{3}} \approx 1 - \frac{x^3}{3}\)
\((1 - x^2)^{\frac{1}{2}} - 1 \approx - \frac{x^2}{2}\)
\(\sin x \approx x\)

Now compute the numerator:

\(e^{x^3} - (1 - x^3)^{\frac{1}{3}} + \left( (1 - x^2)^{\frac{1}{2}} - 1 \right) \sin x\)

\(\approx (1 + x^3) - \left(1 - \frac{x^3}{3}\right) + \left(- \frac{x^2}{2}\right) x\)

\(\approx 1 + x^3 - 1 + \frac{x^3}{3} - \frac{x^3}{2}\)

\(\approx x^3 + \frac{x^3}{3} - \frac{x^3}{2}\)

\(= x^3 \left(1 + \frac{1}{3} - \frac{1}{2}\right)\)

\(= x^3 \left(\frac{6}{6} + \frac{2}{6} - \frac{3}{6}\right)\)

\(= x^3 \left(\frac{5}{6}\right)\)

For the denominator:
\(x \sin^2 x \approx x \cdot x^2 = x^3\)

Thus, the limit becomes:
\(\beta = \lim_{x \to 0} \frac{\frac{5}{6} x^3}{x^3} = \frac{5}{6}\)

Therefore, the value of \(6\beta\) is:

\(6\beta = 6 \times \frac{5}{6} = 5\)

So, the answer is 5.