A) \[1/2\]
B) \[\infty \]
C) 1
D) zero
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{x}}-1}{x} \right)\]as we know that \[{{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+...\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+...1}{x}\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x\left\{ 1+\frac{x}{2!}+\frac{{{x}^{2}}}{3!}+... \right.}{x}\] \[=1\]
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