Q59 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
If the function $f(x) = \left(\dfrac{5x - 8}{8 - 3x}\right)^{\frac{3}{2x - 4}}$, for $x \neq 2$ is continuous at $x = 2$, then the value of $f(2)$ is...
- A$e^{12}$
- Correct answerB$e^6$
- C$e^3$
- D$e^{\frac{3}{2}}$
Solution
Since $f(x)$ is continuous at $x = 2$, $f(2) = \lim_{x \to 2} f(x)$.
$f(2) = \lim_{x \to 2} \left(\dfrac{5x - 8}{8 - 3x}\right)^{\dfrac{3}{2x - 4}}$
As $x \to 2$, the base approaches $1$ and the exponent approaches $\infty$, so this is of the form $1^{\infty}$.
Using the standard limit $\lim_{x \to a} [h(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[h(x) - 1]}$ for $1^{\infty}$ form:
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2x - 4} \left(\dfrac{5x - 8}{8 - 3x} - 1\right)}$
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2(x - 2)} \left(\dfrac{5x - 8 - (8 - 3x)}{8 - 3x}\right)}$
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2(x - 2)} \left(\dfrac{8x - 16}{8 - 3x}\right)}$
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2(x - 2)} \dfrac{8(x - 2)}{8 - 3x}}$
$f(2) = e^{\lim_{x \to 2} \dfrac{12}{8 - 3x}}$
Substituting $x = 2$:
$f(2) = e^{\dfrac{12}{8 - 6}} = e^6$
Answer: $e^6$
$f(2) = \lim_{x \to 2} \left(\dfrac{5x - 8}{8 - 3x}\right)^{\dfrac{3}{2x - 4}}$
As $x \to 2$, the base approaches $1$ and the exponent approaches $\infty$, so this is of the form $1^{\infty}$.
Using the standard limit $\lim_{x \to a} [h(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[h(x) - 1]}$ for $1^{\infty}$ form:
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2x - 4} \left(\dfrac{5x - 8}{8 - 3x} - 1\right)}$
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2(x - 2)} \left(\dfrac{5x - 8 - (8 - 3x)}{8 - 3x}\right)}$
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2(x - 2)} \left(\dfrac{8x - 16}{8 - 3x}\right)}$
$f(2) = e^{\lim_{x \to 2} \dfrac{3}{2(x - 2)} \dfrac{8(x - 2)}{8 - 3x}}$
$f(2) = e^{\lim_{x \to 2} \dfrac{12}{8 - 3x}}$
Substituting $x = 2$:
$f(2) = e^{\dfrac{12}{8 - 6}} = e^6$
Answer: $e^6$
Hence, the correct answer is option: 'B'.