## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$
$$A=\lim_{x\to0}\frac{e^x-1-x}{x^2}$$ Since $\lim_{x\to0}(e^x-1-x)=e^0-1-0=1-1=0$ and $\lim_{x\to0}(x^2)=0^2=0,$ this is an indeterminate form of $\frac{0}{0}$, which can be dealt with by L'Hospital's Rule: $$A=\lim_{x\to0}\frac{(e^x-1-x)'}{(x^2)'}$$ $$A=\lim_{x\to0}\frac{e^x-1}{2x}.$$ $\lim_{x\to0}(e^x-1)=e^0-1=1-1=0$ and $\lim_{x\to0}(2x)=2\times0=0$ again an indeterminate form of $\frac{0}{0}$, so we would apply L'Hospital's Rule one more time: $$A=\lim_{x\to0}\frac{(e^x-1)'}{(2x)'}$$ $$A=\lim_{x\to0}\frac{e^x}{2}$$ $$A=\frac{e^0}{2}$$ $$A=\frac{1}{2}$$