If f(x)=e[x]+x−1[x]+x,x≠01,x=0then,

limx→0+ f(x)=limx→0+ f(0+h) =limh→0 eh−1h=1and limx→0− f(x)=limh→0 f(0−h)=limh→0 e(−h)−h−1(−h)−h=limh→0 e−1−h−1(−1−h)=e−1−1−1=1−1eSince limx→0+ f(x)≠limx→0− f(x)therefore, f (x) is not continuous at x = 0.

If f(x)=e[x]+x−1[x]+x,x≠01,x=0then,

limx→0+ f(x)=limx→0+ f(0+h) =limh→0 eh−1h=1and limx→0− f(x)=limh→0 f(0−h)=limh→0 e(−h)−h−1(−h)−h=limh→0 e−1−h−1(−1−h)=e−1−1−1=1−1eSince limx→0+ f(x)≠limx→0− f(x)therefore, f (x) is not continuous at x = 0.

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If $f (x) = \{\begin{array}{cc} \frac{e^{[x] + x} - 1}{[x] + x} & , x \neq 0 \\ 1 & , x = 0 \end{array}$ then,

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$lim_{x \to 0^{+}} f (x) = - 1$

$lim_{x \to 0^{-}} f (x) = \frac{1}{e} - 1$

$f (x)$ is continuous at x = 0

f (x) is discontinuous at x = 0.

answer is D.

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Detailed Solution

$\begin{array}{l} lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} f (0 + h) \\ = lim_{h \to 0} \frac{e^{h} - 1}{h} = 1 \end{array}$
and $lim_{x \to 0^{-}} f (x) = lim_{h \to 0} f (0 - h)$
$\begin{array}{l} = lim_{h \to 0} \frac{e^{(- h) - h} - 1}{(- h) - h} \\ = lim_{h \to 0} \frac{e^{- 1 - h} - 1}{(- 1 - h)} = \frac{e^{- 1} - 1}{- 1} \\ = 1 - \frac{1}{e} \end{array}$
Since $lim_{x \to 0^{+}} f (x) \neq lim_{x \to 0^{-}} f (x)$
therefore, f (x) is not continuous at x = 0.