Question

If $f(x)=\begin{cases}\frac{\log_{e}(1+\frac{x}{a})-\log_{e}(1-\frac{x}{b})}{x}& if~x\ne0\\k& if~x=0\end{cases}$ is continuous at $x=0$, then value of $k$ is:

• A. 0
• B. 1
• C. $\frac{1}{a}+\frac{1}{b}$
• D. $\frac{1}{a}-\frac{1}{b}$

Hint:

Using Taylor expansion for $\ln(1+u) \approx u - u^2/2 + \dots$ makes these limits very clear: the numerator is $(x/a) - (-x/b) = x(1/a + 1/b)$. Dividing by $x$ gives the result immediately.

Answer 1

The Correct Option is C

For the function $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches 0 must exist and be equal to the value defined at $x=0$, which is $k$.
Step 1: Continuous at $x=0 \implies \lim_{x \to 0} f(x) = f(0)$. Therefore, $k = \lim_{x \to 0} \frac{\log_e(1 + \frac{x}{a}) - \log_e(1 - \frac{x}{b})}{x}$.
Step 2: We use the fundamental limit: $\lim_{u \to 0} \frac{\log_e(1 + u)}{u} = 1$. We rewrite the expression as two separate limits: $k = \lim_{x \to 0} \frac{\log_e(1 + \frac{x}{a})}{x} - \lim_{x \to 0} \frac{\log_e(1 - \frac{x}{b})}{x}$.
Step 3: For the first term, multiply and divide by $a$: $\lim_{x \to 0} \frac{\log_e(1 + \frac{x}{a})}{\frac{x}{a} \cdot a} = \frac{1}{a} \cdot \lim_{\frac{x}{a} \to 0} \frac{\log_e(1 + \frac{x}{a})}{\frac{x}{a}} = \frac{1}{a} \cdot 1 = \frac{1}{a}$. For the second term, multiply and divide by $-b$: $\lim_{x \to 0} \frac{\log_e(1 - \frac{x}{b})}{\frac{x}{-b} \cdot (-b)} = -\frac{1}{b} \cdot \lim_{\frac{-x}{b} \to 0} \frac{\log_e(1 - \frac{x}{b})}{-\frac{x}{b}} = -\frac{1}{b} \cdot 1 = -\frac{1}{b}$.
Step 4: $k = \frac{1}{a} - (-\frac{1}{b}) = \frac{1}{a} + \frac{1}{b}$. The correct value of $k$ is $\frac{1}{a} + \frac{1}{b}$, matching Option (3).