Ok I have this
as lim x --> 0 for
$\displaystyle \frac{sin^2x}{x^2}$
and $\displaystyle sin^2(0)=0$
and $\displaystyle 0^2=0$
so the derivative as lim x-->0
$\displaystyle \frac{2sinxcosx}{2x}$
and i plug in 0 i get 0 so the limit is 0?
So do you only apply the L Hopital Rule When the lim of each denom and num are not both 0 or postive or negative infinity ?
how about for this one ?
lim x-->0
$\displaystyle \frac{e^{-x}}{x}$
I do not see how you got 1 it is my fault because we are taught to take the limit of each num and denom
L'Hôpital's rule is not needed for this limit. You can use it (twice), but that is more work. You should have been taught the following two special trigonometric limits.
$\displaystyle \lim_{x\to0}\frac{\sin x}x=1$
$\displaystyle \lim_{x\to0}\frac{1-\cos x}x=0$
Rewrite, $\displaystyle \frac{e^{-x}}x=\frac1{xe^x}.$ Since $\displaystyle e^x$ is always positive, it should be clear that as $\displaystyle x\to0,\;\frac1{xe^x}$ tends to $\displaystyle \infty$ from the right, and $\displaystyle -\infty$ from the left. This limit does not exist.how about for this one ?
lim x-->0
$\displaystyle \frac{e^{-x}}{x}$