Q58 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
$\lim\limits_{x \to 0}\left[\dfrac{x \cdot \log(1 + 4x)}{\left(e^{4x} - 1\right)^2}\right] = \cdots$
- Correct answerA$\dfrac{1}{4}$
- B$\dfrac{1}{16}$
- C$\dfrac{1}{3}$
- D$\dfrac{1}{9}$
Solution
$\lim\limits_{x \to 0} \left[ \dfrac{x \cdot \log(1 + 4x)}{\left(e^{4x} - 1\right)^2} \right]$
Multiplying and dividing by appropriate powers of $4x$:
$= \lim\limits_{x \to 0} \left[ \dfrac{x \cdot \dfrac{\log(1 + 4x)}{4x} \cdot 4x}{\left(\dfrac{e^{4x} - 1}{4x} \cdot 4x\right)^2} \right]$
$= \lim\limits_{x \to 0} \left[ \dfrac{4x^2 \cdot \dfrac{\log(1 + 4x)}{4x}}{16x^2 \cdot \left(\dfrac{e^{4x} - 1}{4x}\right)^2} \right]$
$= \dfrac{4}{16} \cdot \dfrac{\lim\limits_{x \to 0} \dfrac{\log(1 + 4x)}{4x}}{\left(\lim\limits_{x \to 0} \dfrac{e^{4x} - 1}{4x}\right)^2}$
Using the standard limits $\lim\limits_{u \to 0} \dfrac{\log(1 + u)}{u} = 1$ and $\lim\limits_{u \to 0} \dfrac{e^u - 1}{u} = 1$:
$= \dfrac{1}{4} \cdot \dfrac{1}{(1)^2} = \dfrac{1}{4}$
Answer: $\dfrac{1}{4}$
Multiplying and dividing by appropriate powers of $4x$:
$= \lim\limits_{x \to 0} \left[ \dfrac{x \cdot \dfrac{\log(1 + 4x)}{4x} \cdot 4x}{\left(\dfrac{e^{4x} - 1}{4x} \cdot 4x\right)^2} \right]$
$= \lim\limits_{x \to 0} \left[ \dfrac{4x^2 \cdot \dfrac{\log(1 + 4x)}{4x}}{16x^2 \cdot \left(\dfrac{e^{4x} - 1}{4x}\right)^2} \right]$
$= \dfrac{4}{16} \cdot \dfrac{\lim\limits_{x \to 0} \dfrac{\log(1 + 4x)}{4x}}{\left(\lim\limits_{x \to 0} \dfrac{e^{4x} - 1}{4x}\right)^2}$
Using the standard limits $\lim\limits_{u \to 0} \dfrac{\log(1 + u)}{u} = 1$ and $\lim\limits_{u \to 0} \dfrac{e^u - 1}{u} = 1$:
$= \dfrac{1}{4} \cdot \dfrac{1}{(1)^2} = \dfrac{1}{4}$
Answer: $\dfrac{1}{4}$
Hence, the correct answer is option: 'A'.