Key Concepts and Formulas

• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{F(x)}{G(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{F(x)}{G(x)} = \mathop {\lim }\limits_{x \to c} \frac{F'(x)}{G'(x)}$, provided the latter limit exists.
• Leibniz Integral Rule: For an integral of the form $H(x) = \int_{a(x)}^{b(x)} h(t) dt$, its derivative is $H'(x) = h(b(x)) \cdot b'(x) - h(a(x)) \cdot a'(x)$.
• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^{u(x)} f(t) dt$, then $F'(x) = f(u(x)) \cdot u'(x)$. This is a special case of the Leibniz Integral Rule where the lower limit is a constant.

Step-by-Step Solution

We are given the equation: $\int\limits_6^{f\left( x \right)} {4{t^3}} dt = (x - 2)g(x)$ We need to find $\mathop {\lim }\limits_{x \to 2} g\left( x \right)$.

Step 1: Isolate $g(x)$ and Formulate the Limit To find the limit of $g(x)$, we first express $g(x)$ from the given equation: $g(x) = \frac{\int\limits_6^{f\left( x \right)} {4{t^3}} dt}{x - 2}$ Now, we write the limit we need to evaluate: $\mathop {\lim }\limits_{x \to 2} g\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{\int\limits_6^{f\left( x \right)} {4{t^3}} dt}{x - 2}$ Explanation: This step is a simple algebraic manipulation to get $g(x)$ into a form where we can apply limit techniques.

Step 2: Check for Indeterminate Form Before applying L'Hôpital's Rule, we must confirm that the limit results in an indeterminate form. Let's evaluate the numerator and the denominator as $x \to 2$.

• Numerator: As $x \to 2$, $f(x) \to f(2)$. We are given $f(2) = 6$. So, the numerator becomes: $\int\limits_6^{f\left( 2 \right)} {4{t^3}} dt = \int\limits_6^{6} {4{t^3}} dt$ A definite integral with identical upper and lower limits is always zero. Thus, the numerator approaches $0$.

• Denominator: As $x \to 2$, the denominator $(x - 2)$ approaches $(2 - 2) = 0$.

Since the limit is of the form $\frac{0}{0}$, it is an indeterminate form, and we can apply L'Hôpital's Rule. Explanation: Verifying the indeterminate form is a critical prerequisite for using L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule and Leibniz Integral Rule According to L'Hôpital's Rule, we need to find the derivatives of the numerator and the denominator.

• Derivative of the Denominator: $\frac{d}{dx}(x - 2) = 1$

• Derivative of the Numerator: Let $N(x) = \int\limits_6^{f\left( x \right)} {4{t^3}} dt$. We use the Leibniz Integral Rule (or its special case, the Fundamental Theorem of Calculus Part 1). Here, $h(t) = 4t^3$, the lower limit $a(x) = 6$ (a constant), and the upper limit $b(x) = f(x)$. The derivative is: $N'(x) = h(f(x)) \cdot \frac{d}{dx}(f(x)) - h(6) \cdot \frac{d}{dx}(6)$ $N'(x) = 4(f(x))^3 \cdot f'(x) - 4(6)^3 \cdot 0$ $N'(x) = 4(f(x))^3 \cdot f'(x)$

Now, applying L'Hôpital's Rule: $\mathop {\lim }\limits_{x \to 2} g\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{N'(x)}{1} = \mathop {\lim }\limits_{x \to 2} \frac{4(f(x))^3 \cdot f'(x)}{1}$ Explanation: We apply L'Hôpital's Rule by differentiating both the numerator and the denominator. The Leibniz Integral Rule is essential for differentiating the integral with a variable upper limit.

Step 4: Substitute Given Values and Calculate the Limit Now we can evaluate the limit by substituting $x=2$ into the simplified expression for the derivatives: $\mathop {\lim }\limits_{x \to 2} g\left( x \right) = 4(f(2))^3 \cdot f'(2)$ We are given:

• $f(2) = 6$
• $f'(2) = \frac{1}{48}$

Substitute these values into the expression: $4 \cdot (6)^3 \cdot \frac{1}{48}$ Calculate $6^3$: $6^3 = 216$ Now, compute the final value: $4 \cdot 216 \cdot \frac{1}{48} = \frac{4 \cdot 216}{48}$ We can simplify this calculation: $\frac{4}{48} = \frac{1}{12}$ So, the expression becomes: $216 \cdot \frac{1}{12} = \frac{216}{12}$ Dividing 216 by 12: $\frac{216}{12} = 18$ Thus, the limit is $18$. Explanation: This step involves substituting the given values of $f(2)$ and $f'(2)$ into the derived expression and performing the final arithmetic calculation to find the limit.

Common Mistakes & Tips

• Incorrect application of Leibniz Rule: Ensure that the derivative of the upper limit function ($f'(x)$ in this case) is multiplied correctly, and the derivative of the lower constant limit is zero.
• Forgetting to check for indeterminate form: L'Hôpital's Rule is only applicable to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms. If applied otherwise, the result will be incorrect.
• Algebraic errors in calculation: Double-check the arithmetic, especially when dealing with exponents and fractions.

Summary

The problem requires us to find a limit of a function $g(x)$ which is defined using a definite integral. By rearranging the given equation, we obtain $g(x)$ as a fraction. We observe that the limit of $g(x)$ as $x \to 2$ is of the indeterminate form $\frac{0}{0}$, allowing us to apply L'Hôpital's Rule. Differentiating the numerator requires the use of the Leibniz Integral Rule. After applying these rules and substituting the given values of $f(2)$ and $f'(2)$, we arrive at the final numerical value of the limit.

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