Hi

I need some help on the following questions:

1) Use L'Hopital's rule to obtain:$\displaystyle \lim_{x \to 0} \frac{e^{x}-x-1}{3x^2}$

$\displaystyle f'(x) = \frac{3x^{2}*(e^{x}-1)-(e^{x}-x-1)*6x^2}{9x^4}$

$\displaystyle f'(x) = \frac{3x^{2}e^{x} - 6xe^{x}+3x^2+6x}{9x^4}$

$\displaystyle f'(x) = \frac{3x(xe^{x} - 2e^{x} +3x+2}{9x^4}$

$\displaystyle

f'(x) = \frac{(xe^{x} - 2e^{x} +3x+2}{x^3}$

so if x=0, i get 0 however book's answer says $\displaystyle \frac{1}{6}$

2) Use L'Hopital's rule to obtain: $\displaystyle \lim_{x \to \infty} \frac{x+2}{x^2+2x+1}$

$\displaystyle f'(x) = \frac{1}{2x+2} = \frac{1}{2\infty+2}$

The only problem i am having is what does $\displaystyle \infty$ equal to?

P.S