Key Concepts and Formulas

• Logarithm Properties: The difference of logarithms can be expressed as the logarithm of a quotient: $\log a - \log b = \log \left(\frac{a}{b}\right)$.
• Limit Definition (Implicitly used for L'Hopital's Rule): A limit of a function $f(x)/g(x)$ as $x$ approaches $c$ can be evaluated using L'Hopital's Rule if it results in an indeterminate form (like 0/0 or $\infty/\infty$).
• L'Hopital's Rule: If $\mathop {\lim }\limits_{x \to c} f(x) = 0$ and $\mathop {\lim }\limits_{x \to c} g(x) = 0$, 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.
• Derivative of Logarithm: The derivative of $\log(u)$ with respect to $x$ is $\frac{1}{u} \frac{du}{dx}$.

Step-by-Step Solution

We are asked to find the value of the limit: $k = \mathop {\lim }\limits_{x \to 0} {{\log \left( {3 + x} \right) - \log \left( {3 - x} \right)} \over x}$

Step 1: Check for Indeterminate Form First, let's evaluate the numerator and the denominator as $x$ approaches 0. Numerator: $\log(3+0) - \log(3-0) = \log(3) - \log(3) = 0$. Denominator: $x \to 0$. Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hopital's Rule.

Step 2: Apply L'Hopital's Rule L'Hopital's Rule states that if $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$. Here, $f(x) = \log(3+x) - \log(3-x)$ and $g(x) = x$. We need to find the derivatives of $f(x)$ and $g(x)$.

The derivative of $g(x) = x$ is $g'(x) = 1$.

To find the derivative of $f(x) = \log(3+x) - \log(3-x)$, we differentiate each term: The derivative of $\log(3+x)$ with respect to $x$ is $\frac{1}{3+x} \cdot \frac{d}{dx}(3+x) = \frac{1}{3+x} \cdot 1 = \frac{1}{3+x}$. The derivative of $\log(3-x)$ with respect to $x$ is $\frac{1}{3-x} \cdot \frac{d}{dx}(3-x) = \frac{1}{3-x} \cdot (-1) = -\frac{1}{3-x}$. So, $f'(x) = \frac{1}{3+x} - \left(-\frac{1}{3-x}\right) = \frac{1}{3+x} + \frac{1}{3-x}$.

Now, applying L'Hopital's Rule: $k = \mathop {\lim }\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to 0} \frac{\frac{1}{3+x} + \frac{1}{3-x}}{1}$

Step 3: Evaluate the Limit of the Derivatives Now, substitute $x=0$ into the expression for the derivatives: $k = \frac{\frac{1}{3+0} + \frac{1}{3-0}}{1}$ $k = \frac{\frac{1}{3} + \frac{1}{3}}{1}$ $k = \frac{2}{3}$

Common Mistakes & Tips

• Incorrectly applying L'Hopital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule. Incorrectly applying it will lead to a wrong answer.
• Errors in Differentiation: Carefully differentiate logarithmic functions and use the chain rule correctly, especially when the argument of the logarithm is not simply $x$. Pay close attention to signs, like the derivative of $(3-x)$.
• Algebraic Simplification Errors: After applying L'Hopital's rule, simplify the resulting expression correctly before substituting the limit value.

Summary

The problem requires evaluating a limit that results in an indeterminate form. L'Hopital's Rule is the most direct method. We first verified that the limit is of the form $\frac{0}{0}$ by direct substitution. Then, we applied L'Hopital's Rule by differentiating the numerator and the denominator separately. The derivative of the numerator was found to be $\frac{1}{3+x} + \frac{1}{3-x}$, and the derivative of the denominator was 1. Evaluating the limit of this ratio as $x$ approaches 0 yielded the value $\frac{2}{3}$.

The final answer is $\boxed{\frac{2}{3}}$.