A) 1
B) ?1
C) 0
D) Does not exist
Correct Answer: D
Solution :
\[\underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{0-h}{h+{{h}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{-1}{1+h}=-1\] and \[\underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{h}{h+{{h}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{1}{1+h}=1\] Hence limit does not exist.
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