AP Calculus ABhardmcq1 pt

What is limx0ex1xx2\displaystyle\lim_{x\to 0}\dfrac{e^x-1-x}{x^2}?

A.00
B.11
C.22
D.12\dfrac{1}{2}

Explanation

Core Concept

The form is 00\dfrac{0}{0}. Applying L'Hôpital's rule twice gives ex12xex212\dfrac{e^x-1}{2x}\to\dfrac{e^x}{2}\to\dfrac{1}{2}.

Correct Answer

D12\dfrac{1}{2}

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