MHTCET 17 Apr 2026 Shift 1 Paper

Concept:A function is continuous at $x=0$ if $\lim_{x \to 0} f(x) = f(0)$.If the limit does not exist or is not equal to $f(0)$, the function is discontinuous at $x=0$.Explanation:Check the limit for each option and compare it with the given value of $f(0)$.For Option A:$\lim_{x \to 0} (1+x)^{\frac{2}{x}} = e^2$ and $f(0) = e^2$.So Option A is continuous at $x=0$.For Option B:$\lim_{x \to 0} (\sin x - \cos x) = \sin 0 - \cos 0 = -1$ and $f(0) = -1$.So Option B is continuous at $x=0$.For Option C:As $x \to 0^+$, $\frac{1}{x} \to \infty$, so $e^{\frac{1}{x}} \to \infty$.Thus, $\lim_{x \to 0^+} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1} = 1$.As $x \to 0^-$, $\frac{1}{x} \to -\infty$, so $e^{\frac{1}{x}} \to 0$.Thus, $\lim_{x \to 0^-} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1} = \frac{0-1}{0+1} = -1$.Since right-hand limit and left-hand limit are not equal, $\lim_{x \to 0} f(x)$ does not exist.Hence Option C is discontinuous at $x=0$.For Option D:Using standard limits, $e^{5x}-e^{2x} \approx 5x-2x = 3x$ and $\sin 3x \approx 3x$.So, $\lim_{x \to 0} \frac{e^{5x}-e^{2x}}{\sin 3x} = 1$ and $f(0) = 1$.So Option D is continuous at $x=0$.Answer:Option C