Key Concepts and Formulas

1. L'Hôpital's Rule: This rule is used to evaluate limits of indeterminate forms of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. If $\lim_{x \to c} \frac{F(x)}{G(x)}$ is an indeterminate form, then $\lim_{x \to c} \frac{F(x)}{G(x)} = \lim_{x \to c} \frac{F'(x)}{G'(x)}$, provided the latter limit exists.
2. Leibniz Integral Rule (Fundamental Theorem of Calculus, Part 1, for variable limits): This rule states that if $g(t)$ is a continuous function and $a(x)$, $b(x)$ are differentiable functions, then: $\frac{d}{dx} \int_{a(x)}^{b(x)} g(t) dt = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x)$
3. Properties of Definite Integrals: An integral with identical upper and lower limits is always zero: $\int_a^a g(t) dt = 0$.

Step-by-Step Solution

Step 1: Analyze the Limit and Identify Indeterminate Form We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt}$ To apply L'Hôpital's Rule, we first rewrite the expression as a fraction: $L = \mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$ Now, we examine the behavior of the numerator and the denominator as $x \to 2$:

• Numerator: As $x \to 2$, since $f(x)$ is differentiable, it is also continuous. Therefore, $f(x) \to f(2)$. We are given $f(2) = 6$. So, the numerator approaches $\int_6^{f(2)} 4t^3 dt = \int_6^6 4t^3 dt$. Since the upper and lower limits are the same, the value of the integral is $0$.
• Denominator: As $x \to 2$, the denominator $x-2$ approaches $2-2 = 0$.

Since both the numerator and the denominator approach $0$ as $x \to 2$, the limit is of the indeterminate form $\frac{0}{0}$. This indicates that L'Hôpital's Rule can be applied.

Step 2: Apply L'Hôpital's Rule by Differentiating Numerator and Denominator We differentiate the numerator and the denominator with respect to $x$.

• Differentiating the Numerator: We need to find $\frac{d}{dx} \left( \int\limits_6^{f\left( x \right)} {4{t^3}dt} \right)$. We use the Leibniz Integral Rule. Here, $g(t) = 4t^3$, the lower limit $a(x) = 6$ (a constant), and the upper limit $b(x) = f(x)$ (a differentiable function). Applying the rule: $\frac{d}{dx} \left( \int\limits_6^{f\left( x \right)} {4{t^3}dt} \right) = g(f(x)) \cdot f'(x) - g(6) \cdot \frac{d}{dx}(6)$ Substituting $g(t) = 4t^3$: $= \left[ 4(f(x))^3 \right] \cdot f'(x) - \left[ 4(6)^3 \right] \cdot (0)$ The derivative of the constant $6$ is $0$, so the second term vanishes. The derivative of the numerator is $4(f(x))^3 f'(x)$.

• Differentiating the Denominator: We need to find $\frac{d}{dx} (x-2)$. $\frac{d}{dx} (x-2) = 1$

Step 3: Substitute the Derivatives back into the Limit Expression Applying L'Hôpital's Rule, the limit becomes: $L = \mathop {\lim }\limits_{x \to 2} {{4{{\left( {f\left( x \right)} \right)}^3}f'\left( x \right)} \over 1}$ $L = \mathop {\lim }\limits_{x \to 2} 4{\left( {f\left( x \right)} \right)^3}f'\left( x \right)$

Step 4: Evaluate the Limit Using Given Information Since $f(x)$ is differentiable, it is continuous, and $f'(x)$ is also continuous. Therefore, we can directly substitute $x=2$ into the expression: $L = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$ We are given $f(2) = 6$ and $f'(2) = \frac{1}{48}$. Substituting these values: $L = 4 \times \left( 6 \right)^3 \times \left( \frac{1}{48} \right)$ Calculate $6^3$: $6^3 = 6 \times 6 \times 6 = 216$ Now substitute this back: $L = 4 \times 216 \times \frac{1}{48}$ $L = \frac{4 \times 216}{48}$ We can simplify the fraction: $L = \frac{216}{12}$ $L = 18$

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed of an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule.
• Errors in differentiating the integral: Carefully use the Leibniz Integral Rule, especially when the limits of integration are functions of $x$. Remember to multiply by the derivative of the limit function.
• Algebraic simplification errors: Double-check all arithmetic and algebraic manipulations, especially when dealing with fractions and exponents.

Summary

The problem involves evaluating a limit of a definite integral. We first identified that the limit is of the indeterminate form $\frac{0}{0}$ by examining the numerator and denominator as $x \to 2$. This allowed us to apply L'Hôpital's Rule. Differentiating the numerator required the use of the Leibniz Integral Rule, and differentiating the denominator was straightforward. After applying L'Hôpital's Rule and substituting the given values of $f(2)$ and $f'(2)$, we performed algebraic simplification to arrive at the final answer.

The final answer is $\boxed{18}$.