Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\mathop {\lim }\limits_{x \to c} f(x)$ exists.
  3. $\mathop {\lim }\limits_{x \to c} f(x) = f(c)$.
• Standard Limit Form: The fundamental limit form $\mathop {\lim }\limits_{y \to 0} \frac{\ln(1+y)}{y} = 1$ is crucial for evaluating limits involving logarithmic functions.
• Logarithm Properties: $\ln(a/b) = \ln(a) - \ln(b)$.

Step-by-Step Solution

Step 1: Understand the Condition for Continuity The problem states that the function $f(x)$ is continuous on the interval $\left( { - {1 \over 3},{1 \over 3}} \right)$. For the function to be continuous at $x=0$, the limit of the function as $x$ approaches 0 must be equal to the value of the function at $x=0$. That is, $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$. We are given $f(0) = k$, so we need to find the limit of $f(x)$ as $x \to 0$.

Step 2: Express $f(x)$ using Logarithm Properties For $x \ne 0$, the function is defined as $f(x) = \frac{1}{x}\log_e\left(\frac{1+3x}{1-2x}\right)$. Using the property of logarithms $\log_e(a/b) = \log_e(a) - \log_e(b)$, we can rewrite $f(x)$ as: $f(x) = \frac{1}{x} \left( \log_e(1+3x) - \log_e(1-2x) \right)$ $f(x) = \frac{\log_e(1+3x)}{x} - \frac{\log_e(1-2x)}{x}$

Step 3: Evaluate the Limit of $f(x)$ as $x \to 0$ Now we need to find $\mathop {\lim }\limits_{x \to 0} f(x)$. We can split this into two separate limits: $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \left( \frac{\log_e(1+3x)}{x} - \frac{\log_e(1-2x)}{x} \right)$ $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} \frac{\log_e(1+3x)}{x} - \mathop {\lim }\limits_{x \to 0} \frac{\log_e(1-2x)}{x}$

Step 4: Apply the Standard Limit Form To evaluate each of these limits, we will use the standard limit form $\mathop {\lim }\limits_{y \to 0} \frac{\ln(1+y)}{y} = 1$.

For the first limit, $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1+3x)}{x}$: Let $y = 3x$. As $x \to 0$, $y \to 0$. We can rewrite the expression by multiplying the numerator and denominator by 3: $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1+3x)}{x} = \mathop {\lim }\limits_{x \to 0} 3 \cdot \frac{\log_e(1+3x)}{3x}$ $= 3 \cdot \mathop {\lim }\limits_{y \to 0} \frac{\log_e(1+y)}{y} = 3 \cdot 1 = 3$

For the second limit, $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1-2x)}{x}$: Let $z = -2x$. As $x \to 0$, $z \to 0$. We can rewrite the expression by multiplying the numerator and denominator by -2: $\mathop {\lim }\limits_{x \to 0} \frac{\log_e(1-2x)}{x} = \mathop {\lim }\limits_{x \to 0} (-2) \cdot \frac{\log_e(1-2x)}{-2x}$ $= -2 \cdot \mathop {\lim }\limits_{z \to 0} \frac{\log_e(1+z)}{z} = -2 \cdot 1 = -2$

Step 5: Combine the Limits to Find the Value of $k$ Now, substitute the values of the individual limits back into the expression for $\mathop {\lim }\limits_{x \to 0} f(x)$: $\mathop {\lim }\limits_{x \to 0} f(x) = 3 - (-2)$ $\mathop {\lim }\limits_{x \to 0} f(x) = 3 + 2 = 5$

Since the function is continuous at $x=0$, we have $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$. Therefore, $k = 5$.

Common Mistakes & Tips

• Incorrectly applying the standard limit: Ensure that the term inside the logarithm is matched with the denominator. For example, $\log_e(1+3x)$ requires a $3x$ in the denominator.
• Sign errors: Be careful with negative signs, especially when manipulating the terms to fit the standard limit form, as seen with the $\log_e(1-2x)$ term.
• Forgetting the continuity condition: Remember that continuity at a point means the limit must equal the function's value at that point.

Summary

To find the value of $k$ that makes the function $f(x)$ continuous at $x=0$, we used the definition of continuity: $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$. We first simplified the expression for $f(x)$ using logarithm properties. Then, we evaluated the limit as $x \to 0$ by breaking it into two parts and applying the standard limit form $\mathop {\lim }\limits_{y \to 0} \frac{\ln(1+y)}{y} = 1$. After calculating the limit to be 5, we equated it to $f(0) = k$, thus finding $k=5$.

The final answer is \boxed{5}.