My last pretty straight forward, How about these two.
lim (1 +3h)^1/h
h->0
or
lim 3^-x
x->+infin
do you know what L'Hopital's rule says? remember, $\displaystyle \lim_{x \to a} e^{f(x)} = e^{\lim_{x \to a}f(x)}$
first of all, 1^0 is 1, not zero. secondly, y would tend to infinity as h tends to 0. thirdly, if you are changing h, you would have to change ALL the h's, meaning your new limit would be $\displaystyle \lim_{y \to \infty} \left( 1 + \frac 3y \right)^y$With that question I did y = 1/h as h->0, y->0 therefore the lim (1+3h)^0 is 0
um, yeah. we have a constant numerator and the denominator is growing without bound.As for the 2nd one, is that implying that the answer is 0?