Which of the following functions is discontin | vedclass

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(C) function $f(x)$ is continuous at $x=0$ if $\lim_{x 	o 0} f(x) = f(0)$.$(A)$ $\lim_{x 	o 0} (1+x)^{\frac{2}{x}} = e^{\lim_{x 	o 0} \frac{2}{x} \

Vedclass · Accepted Answer

$f(x) = \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1}$ for $x 
eq 0$, $f(0) = -1$

Vedclass · Answer

(C) function $f(x)$ is continuous at $x=0$ if $\lim_{x 	o 0} f(x) = f(0)$.$(A)$ $\lim_{x 	o 0} (1+x)^{\frac{2}{x}} = e^{\lim_{x 	o 0} \frac{2}{x} \cdot x} = e^2 = f(0)$. Continuous.$(B)$ $\lim_{x 	o 0} (\sin x - \cos x) = 0 - 1 = -1 = f(0)$. Continuous.$(C)$ For $x 	o 0^+$, $\lim_{x 	o 0^+} \frac{e^{1/x}(1 - e^{-1/x})}{e^{1/x}(1 + e^{-1/x})} = 1$. For $x 	o 0^-$, $\lim_{x 	o 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$. Since $	ext{LHL} 
eq 	ext{RHL}$, the limit does not exist. Discontinuous.$(D)$ $\lim_{x 	o 0} \frac{e^{5x} - e^{2x}}{\sin 3x} = \lim_{x 	o 0} \frac{\frac{e^{5x}-1}{x} - \frac{e^{2x}-1}{x}}{\frac{\sin 3x}{3x} \cdot 3} = \frac{5 - 2}{3} = 1 = f(0)$. Continuous.

Which of the following functions is discontinuous at $x = 0$?

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