A) 1/2
B) \[\infty \]
C) 1
D) 0
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\text{lim}}}\,\frac{{{e}^{x}}-1}{x}=\underset{x\to 0}{\mathop{\text{lim}}}\,\frac{\left( 1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+...-1 \right)}{x}=1\].
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