\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \left( {1 + x + \frac{{{x^2}}}{2}} \right)}}{{{x^3}}} =\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \left( {1 + x + \fra

1.	\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \left( {1 + x + \frac{{{x^2}}}{2}} \right)}}{{{x^3}}} =\)
A.	0
B.	\(\frac{1}{6}\)
C.	\(\frac{1}{3}\)
D.	1
Answer» C. \(\frac{1}{3}\)

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\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \left( {1 + x + \frac{{{x^2}}}{2}} \right)}}{{{x^3}}} =\)

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