1. ## Two limit problems

My last pretty straight forward, How about these two.

lim (1 +3h)^1/h
h->0
or

lim 3^-x
x->+infin

2. Originally Posted by Meeklo Braca
That one is pretty straight forward, How about these two.

lim (1 +3h)^1/h
h->0
Note that $\lim_{h \to 0} (1 + 3h)^{1/h} = \lim_{h \to 0} e^{\ln (1 + 3h)^{1/h}} = \lim_{h \to 0}e^{\frac {\ln (1 + 3h)}h}$ Now, apply L'Hopital's rule.

lim 3^-x
x->+infin
recall that $3^{-x} = \frac 1{3^x}$

3. I have no idea how to apply L'Hopital's rule to the first question.

With that question I did y = 1/h as h->0, y->0 therefore the lim (1+3h)^0 is 0

As for the 2nd one, is that implying that the answer is 0?

4. Originally Posted by Meeklo Braca
I have no idea how to apply L'Hopital's rule to the first question.
do you know what L'Hopital's rule says? remember, $\lim_{x \to a} e^{f(x)} = e^{\lim_{x \to a}f(x)}$

With that question I did y = 1/h as h->0, y->0 therefore the lim (1+3h)^0 is 0
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 $\lim_{y \to \infty} \left( 1 + \frac 3y \right)^y$

As for the 2nd one, is that implying that the answer is 0?
um, yeah. we have a constant numerator and the denominator is growing without bound.