Key Concepts and Formulas

1. L'Hôpital's Rule: For limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, if $\lim_{x \to c} \frac{F(x)}{G(x)}$ is indeterminate, 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. Fundamental Theorem of Calculus (Part II): If $F(t)$ is an antiderivative of a continuous function $f(t)$, then $\int_a^b f(t) dt = F(b) - F(a)$.
3. Chain Rule for Differentiation: If $y = g(u)$ and $u = h(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Step-by-Step Solution

We are asked to find the limit: $L = \mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$

Step 1: Evaluate the definite integral. First, let's find the indefinite integral of $2t$. The antiderivative of $2t$ is $t^2$. So, $\int {2t dt} = t^2 + C$. Now, we evaluate the definite integral: $\int\limits_6^{f\left( x \right)} {2tdt} = \left[ t^2 \right]_6^{f\left( x \right)} = (f(x))^2 - 6^2 = (f(x))^2 - 36$ The limit can be rewritten as: $L = \mathop {\lim }\limits_{x \to 2} \frac{(f(x))^2 - 36}{x - 2}$

Step 2: Check for indeterminate form. As $x \to 2$, we are given that $f(2) = 6$. The numerator approaches $(f(2))^2 - 36 = 6^2 - 36 = 36 - 36 = 0$. The denominator approaches $2 - 2 = 0$. Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule. L'Hôpital's Rule states that $\lim_{x \to c} \frac{F(x)}{G(x)} = \lim_{x \to c} \frac{F'(x)}{G'(x)}$. Here, $F(x) = (f(x))^2 - 36$ and $G(x) = x - 2$.

We need to find the derivatives of the numerator and the denominator with respect to $x$. The derivative of the denominator is: $G'(x) = \frac{d}{dx}(x - 2) = 1$

The derivative of the numerator requires the chain rule: $F'(x) = \frac{d}{dx}((f(x))^2 - 36)$ $F'(x) = \frac{d}{dx}((f(x))^2) - \frac{d}{dx}(36)$ Using the chain rule, $\frac{d}{dx}((f(x))^2) = 2 f(x) \cdot f'(x)$. The derivative of a constant is 0. So, $F'(x) = 2 f(x) f'(x)$.

Now, apply L'Hôpital's Rule to the limit: $L = \mathop {\lim }\limits_{x \to 2} \frac{2 f(x) f'(x)}{1}$

Step 4: Evaluate the limit of the derivatives. Substitute $x = 2$ into the expression: $L = 2 f(2) f'(2)$ We are given that $f(2) = 6$. $L = 2 (6) f'(2)$ $L = 12 f'(2)$

Step 5: Alternative approach using Fundamental Theorem of Calculus and limit definition of derivative. Let $H(x) = \int_6^{f(x)} 2t \, dt$. By the Fundamental Theorem of Calculus Part I, if we let $G(u) = \int_6^u 2t \, dt$, then $G'(u) = 2u$. Using the chain rule, the derivative of $H(x)$ is $H'(x) = G'(f(x)) \cdot f'(x)$. Substituting $G'(u) = 2u$, we get $H'(x) = 2 f(x) f'(x)$.

The limit we need to evaluate is: $L = \mathop {\lim }\limits_{x \to 2} \frac{H(x)}{x - 2}$ This is the definition of the derivative of $H(x)$ at $x=2$, provided $H(2)=0$. Let's check $H(2)$: $H(2) = \int_6^{f(2)} 2t \, dt$. Since $f(2) = 6$, we have: $H(2) = \int_6^6 2t \, dt = 0$.

Thus, the limit is the definition of $H'(2)$: $L = H'(2)$ We found $H'(x) = 2 f(x) f'(x)$. So, $H'(2) = 2 f(2) f'(2)$. Given $f(2) = 6$, we have: $H'(2) = 2 (6) f'(2) = 12 f'(2)$.

Therefore, $L = 12 f'(2)$.

Common Mistakes & Tips

• Incorrect application of the Fundamental Theorem of Calculus: Ensure you correctly apply the chain rule when differentiating an integral whose upper limit is a function of the integration variable.
• Forgetting to check the indeterminate form: Before applying L'Hôpital's Rule, always verify that the limit results in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$). Otherwise, the rule is not applicable.
• Algebraic errors in differentiation: Be careful when differentiating composite functions, especially when squaring a function of $x$ like $(f(x))^2$.

Summary

The problem requires evaluating a limit involving a definite integral. We first evaluated the definite integral to simplify the numerator. Upon observing that the limit is of the indeterminate form $\frac{0}{0}$ as $x \to 2$, we applied L'Hôpital's Rule. Differentiating the numerator using the chain rule and the denominator, and then evaluating the resulting limit at $x=2$ using the given information $f(2)=6$, we arrived at the solution. An alternative method using the definition of the derivative also confirms the result.

The final answer is $\boxed{12f'(2)}$.