Q59 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
Which of the following function is discontinuous at $x = 0$ ?
- A$f(x) = (1 + x)^{\frac{2}{x}},$ for $x \neq 0$
$= e^2,$ for $x = 0$ - B$f(x) = \sin x - \cos x,$ for $x \neq 0$
$= -1,$ for $x = 0$ - Correct answerC$f(x) = \dfrac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1},$ for $x \neq 0$
$= -1,$ for $x = 0$ - D$f(x) = \dfrac{e^{5x} - e^{2x}}{\sin 3x},$ for $x \neq 0$
$= 1,$ for $x = 0$
Solution
Let us check the continuity of each function at $x = 0$ by evaluating the limits.
For the first function:
$\lim_{x \to 0} (1 + x)^{\dfrac{2}{x}} = e^{\lim_{x \to 0} \dfrac{2}{x} \ln(1+x)} = e^2 = f(0)$
The function is continuous at $x = 0$.
For the second function:
$\lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = -1 = f(0)$
The function is continuous at $x = 0$.
For the third function:
Right-hand limit (RHL):
$\lim_{x \to 0^+} \dfrac{e^{\dfrac{1}{x}} - 1}{e^{\dfrac{1}{x}} + 1} = \lim_{x \to 0^+} \dfrac{1 - e^{-\dfrac{1}{x}}}{1 + e^{-\dfrac{1}{x}}} = \dfrac{1 - 0}{1 + 0} = 1$
Left-hand limit (LHL):
$\lim_{x \to 0^-} \dfrac{e^{\dfrac{1}{x}} - 1}{e^{\dfrac{1}{x}} + 1} = \dfrac{0 - 1}{0 + 1} = -1$
Since LHL $\neq$ RHL, the limit does not exist. The function is discontinuous at $x = 0$.
For the fourth function:
$\lim_{x \to 0} \dfrac{e^{5x} - e^{2x}}{\sin 3x} = \lim_{x \to 0} e^{2x} \left( \dfrac{e^{3x} - 1}{3x} \right) \left( \dfrac{3x}{\sin 3x} \right) = 1 \times 1 \times 1 = 1 = f(0)$
The function is continuous at $x = 0$.
Answer: $f(x) = \dfrac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1},$ for $x \neq 0$
$= -1,$ for $x = 0$
For the first function:
$\lim_{x \to 0} (1 + x)^{\dfrac{2}{x}} = e^{\lim_{x \to 0} \dfrac{2}{x} \ln(1+x)} = e^2 = f(0)$
The function is continuous at $x = 0$.
For the second function:
$\lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = -1 = f(0)$
The function is continuous at $x = 0$.
For the third function:
Right-hand limit (RHL):
$\lim_{x \to 0^+} \dfrac{e^{\dfrac{1}{x}} - 1}{e^{\dfrac{1}{x}} + 1} = \lim_{x \to 0^+} \dfrac{1 - e^{-\dfrac{1}{x}}}{1 + e^{-\dfrac{1}{x}}} = \dfrac{1 - 0}{1 + 0} = 1$
Left-hand limit (LHL):
$\lim_{x \to 0^-} \dfrac{e^{\dfrac{1}{x}} - 1}{e^{\dfrac{1}{x}} + 1} = \dfrac{0 - 1}{0 + 1} = -1$
Since LHL $\neq$ RHL, the limit does not exist. The function is discontinuous at $x = 0$.
For the fourth function:
$\lim_{x \to 0} \dfrac{e^{5x} - e^{2x}}{\sin 3x} = \lim_{x \to 0} e^{2x} \left( \dfrac{e^{3x} - 1}{3x} \right) \left( \dfrac{3x}{\sin 3x} \right) = 1 \times 1 \times 1 = 1 = f(0)$
The function is continuous at $x = 0$.
Answer: $f(x) = \dfrac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1},$ for $x \neq 0$
$= -1,$ for $x = 0$
Hence, the correct answer is option: 'C'.