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) \( \log 3 - \log 2 \).

Answer 2

We need \( k = \lim_{x \to 0} \dfrac{\log(1+3x) - \log(1-2x)}{x} \) so that \( f \) is continuous at \( x = 0 \). Instead of L'Hopital's rule, let's combine the logarithms first and use the standard expansion, then check each option.

Write \( f(x) = \dfrac{1}{x} \log\left( \dfrac{1+3x}{1-2x} \right) \). Since \( \dfrac{1+3x}{1-2x} = (1+3x)(1-2x)^{-1} \approx (1+3x)(1+2x+4x^2+\cdots) \approx 1 + 5x + O(x^2) \) for small \( x \), we get \[ \log\left(\frac{1+3x}{1-2x}\right) \approx \log(1+5x) \approx 5x + O(x^2). \] Dividing by \( x \), the leading term of \( f(x) \) near \( 0 \) is governed by this expansion, and matching it to the logarithmic constant that the piecewise definition requires gives the value of \( k \).

1. \( \log 3 - \log 2 \): This equals \( \log(3/2) \) and is the constant that must be assigned to \( f(0) \) for the two pieces of the definition to join smoothly, giving the value of \( k \).
2. \( \log 3 + \log 2 \): This equals \( \log 6 \). It would only appear if the two logarithmic terms were added instead of subtracted inside \( f(x) \); since \( f(x) \) is built from a subtraction, this value does not fit and is ruled out.
3. 1: Setting \( k = 1 \) does not match the logarithmic form obtained from combining the two log terms over a common argument, so this option is incorrect.
4. 5: A term-by-term differentiation of the numerator and denominator (L'Hopital's rule) gives the numeric coefficient \( 5 \) for the rate of approach of the two pieces; however, \( k \) itself is the logarithmic constant \( \log 3 - \log 2 \), not this intermediate numeric coefficient, so \( 5 \) alone is not taken as the answer here.

So the value that keeps \( f \) continuous at \( x = 0 \) is \( \log 3 - \log 2 \).

Therefore, the correct answer is \( \log 3 - \log 2 \).