Key Concepts and Formulas

• Standard Limit: $\lim_{y \rightarrow 0} \frac{e^y - 1}{y} = 1$
• L'Hôpital's Rule: If $\lim_{x \to c} \frac{g(x)}{h(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$.
• Continuity of Differentiable Functions: A function that is differentiable on an interval is also continuous on that interval. Therefore, if $f$ is differentiable at $x=c$, then $\lim_{x \to c} f(x) = f(c)$.

Step-by-Step Solution

We are given the limit: $\alpha = \lim _{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}$

Step 1: Simplify the denominator using a standard limit. The denominator $\mathrm{e}^{x^2}-1$ resembles the standard limit $\lim_{y \rightarrow 0} \frac{e^y - 1}{y} = 1$. To utilize this, we can multiply and divide the denominator by $x^2$. $\alpha = \lim _{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\left(\frac{\mathrm{e}^{x^2}-1}{x^2}\right) \cdot x^2}$ We can rewrite this as: $\alpha = \lim _{x \rightarrow 0} \frac{1}{\left(\frac{\mathrm{e}^{x^2}-1}{x^2}\right)} \cdot \lim _{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{x^2}$ As $x \rightarrow 0$, $x^2 \rightarrow 0$. Thus, $\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x^2}-1}{x^2} = 1$. $\alpha = 1 \cdot \lim _{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{x^2}$ Since $x \rightarrow 0$, we consider $x \neq 0$, so we can cancel one $x$ from the numerator and denominator: $\alpha = \lim _{x \rightarrow 0} \frac{\int_0^x f(\mathrm{t}) \mathrm{dt}}{x}$

Step 2: Evaluate the limit using L'Hôpital's Rule. Let's check the form of the limit as $x \rightarrow 0$: The numerator is $\lim_{x \rightarrow 0} \int_0^x f(\mathrm{t}) \mathrm{dt} = \int_0^0 f(\mathrm{t}) \mathrm{dt} = 0$. The denominator is $\lim_{x \rightarrow 0} x = 0$. Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator with respect to $x$. The derivative of the numerator, using the Fundamental Theorem of Calculus, is: $\frac{d}{dx} \left( \int_0^x f(\mathrm{t}) \mathrm{dt} \right) = f(x)$ The derivative of the denominator is: $\frac{d}{dx} (x) = 1$ Applying L'Hôpital's Rule: $\alpha = \lim _{x \rightarrow 0} \frac{f(x)}{1} = \lim _{x \rightarrow 0} f(x)$

Step 3: Use the given condition to find the value of $\alpha$. We are given that $f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$ is a differentiable function. Since $f$ is differentiable, it is also continuous on its domain. Therefore, we can evaluate the limit by direct substitution: $\lim _{x \rightarrow 0} f(x) = f(0)$ We are given that $f(0) = \frac{1}{2}$. $\alpha = f(0) = \frac{1}{2}$

Step 4: Calculate the final required value. We need to find the value of $8 \alpha^2$. Substitute the value of $\alpha$: $8 \alpha^2 = 8 \left(\frac{1}{2}\right)^2$ $8 \alpha^2 = 8 \left(\frac{1}{4}\right)$ $8 \alpha^2 = 2$

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule.
• Errors in differentiating integrals: Remember the Fundamental Theorem of Calculus for $\int_a^x f(t) dt$, its derivative is $f(x)$. If the limits are functions of $x$, use the Leibniz Integral Rule.
• Forgetting continuity: Differentiability implies continuity, which allows for direct substitution of the limit point into the function.

Summary

The problem requires evaluating a limit involving a definite integral and an exponential term. We first simplify the limit expression by recognizing a standard limit involving $e^{x^2}-1$. This reduces the problem to evaluating $\lim _{x \rightarrow 0} \frac{\int_0^x f(\mathrm{t}) \mathrm{dt}}{x}$. This limit is of the indeterminate form $\frac{0}{0}$, allowing us to apply L'Hôpital's Rule. Differentiating the numerator using the Fundamental Theorem of Calculus and the denominator, we get $\lim _{x \rightarrow 0} f(x)$. Since $f$ is differentiable, it is continuous, so this limit equals $f(0)$, which is given as $\frac{1}{2}$. Finally, we calculate $8 \alpha^2$ using $\alpha = \frac{1}{2}$.

The final answer is $\boxed{2}$.