Question

limit problem

Answer 1

evaluate using l'hopital\[\lim_{x \rightarrow 0} (e^x-1-x)/x^2\]

Answer 2

L'Hôpital says
\[ \large \lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)} \]
where in your case
\[ \large f(x)=e^x-1-x \]
\[ \large g(x)=x^2 \]

Answer 3

I can apply L'hopital more than one to the same problem, correct?

Answer 4

after differentiating you get
lim (e^x -1)/2x

which is still 0/0 so differentiate again
lim e^x / 2 = 1/2

Answer 5

yes

Answer 6

thanks, i appreciate it