A) 0
B) \[\frac{1}{2}\]
C) \[e-2\]
D) None of these
Correct Answer: D
Solution :
[d] Let \[[a]=n,\]then \[\underset{x\to {{n}^{-}}}{\mathop{\lim }}\,\frac{{{e}^{\{x\}}}-\{x\}-1}{{{\{x\}}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{\{n-h\}}}-\{n-h\}-1}{{{\{n-h\}}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{1-h}}-(1-h)-1}{{{(1-h)}^{2}}}\] \[=\,\,\,e-2\] and \[\underset{x\to {{n}^{+}}}{\mathop{\lim }}\,\frac{{{e}^{\{x\}}}-\{x\}-1}{{{\{x\}}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{\{n+h\}}}-\{n+h\}-1}{{{\{n+h\}}^{2}}}\] \[=\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{h}}-h-1}{{{h}^{2}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1+h+\frac{{{h}^{2}}}{2!}+\frac{{{h}^{3}}}{3!}+...-h-1}{{{h}^{2}}}=\frac{1}{2}\] \[\therefore \] Limit does not exist.
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