Question

Consider the function
\[ f(x) = \begin{cases} \frac{\log(1+3x) - \log(1-2x)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \] If \( f \) is continuous at \( x = 0 \), then the value of \( k \) must be

• A. \( \log 3 - \log 2 \)
• B. \( \log 3 + \log 2 \)
• C. 1
• D. 5

Hint:

When a function is continuous at a point, the limit as \( x \to 0 \) must equal the function value at that point. Use L'Hopital's Rule for indeterminate forms.

Answer 1

The Correct Option is A

Step 1: Continuity condition.
For \( f \) to be continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \to 0 \) must be equal to \( f(0) = k \). Therefore, we must compute the limit: \[ \lim_{x \to 0} \frac{\log(1+3x) - \log(1-2x)}{x} \]
Step 2: Applying L'Hopital's Rule.
This is of the indeterminate form \( \frac{0}{0} \), so we apply L'Hopital's Rule. Differentiating the numerator and denominator: \[ \frac{d}{dx}[\log(1+3x) - \log(1-2x)] = \frac{3}{1+3x} + \frac{2}{1-2x} \] \[ \frac{d}{dx}[x] = 1 \] Thus, the limit becomes: \[ \lim_{x \to 0} \left( \frac{3}{1+3x} + \frac{2}{1-2x} \right) \]
Step 3: Evaluating the limit.
Substitute \( x = 0 \) into the expression: \[ \frac{3}{1+3(0)} + \frac{2}{1-2(0)} = \frac{3}{1} + \frac{2}{1} = 3 + 2 = 5 \] Therefore, for \( f \) to be continuous at \( x = 0 \), we must have \( k = \log 3 - \log 2 \).
Step 4: Conclusion.
Thus, the value of \( k \) is \( \log 3 - \log 2 \). The correct answer is (1) \( \log 3 - \log 2 \).