Question

The function $f$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x) = \left\{ \begin{array}{ll} \frac{1}{x} \log\left(\frac{1+3x}{1-2x}\right) & , x \neq 0 \\> k & , x = 0 \end{array} \right.$ is continuous at $x = 0$, then $k$ is

• A. 6
• B. 1
• C. 5
• D. -5

Hint:

Recall that for a function to be continuous at a point $x=a$, the limit of the function as $x$ approaches $a$ must be equal to the function's value at $a$. Also, remember the standard limit $\lim_{x \to 0} \frac{\log(1+ax)}{ax} = 1$.

Answer 1

The Correct Option is A

Step 1: For $f(x)$ to be continuous at $x=0$, the limit of the function as $x$ approaches $0$ must be equal to the function's value at $x=0$. That is, $f(0) = \lim_{x \to 0} f(x)$.
Step 2: Given $f(0) = k$, we need to evaluate the limit:\n\[k = \lim_{x \to 0} \frac{1}{x} \log\left(\frac{1+3x}{1-2x}\right)\]
Step 3: Use the logarithm property $\log\left(\frac{A}{B}\right) = \log(A) - \log(B)$ to separate the terms.\n\[k = \lim_{x \to 0} \left(\frac{1}{x} \log(1+3x) - \frac{1}{x} \log(1-2x)\right)\]
Step 4: Rearrange the terms to use the standard limit $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$. We multiply and divide by the coefficients of $x$ inside the logarithm.\n\[k = \lim_{x \to 0} \left(\frac{3 \log(1+3x)}{3x} + \frac{2 \log(1-2x)}{-2x}\right)\]
Step 5: Evaluate the limit. As $x \to 0$, $3x \to 0$ and $-2x \to 0$. Therefore, applying the standard limit:\n\[k = 3(1) + 2(1) = 5\]