solving for lim?

Question

the question im trying to figure out is lim (below it: x->0) (1+3x)^1/x . i so far was able to get 3 /(x+3x^2) but that is undefined. what am i doing thats worng?

Answer 1

The quick answer is e^3 (that's what Mathematica is telling me). Let me see if I can figure out why.

Edit:

Do a power series expansion around x=0. Mathematica tells me this is:

e^3(1+9x/2+153x^2/8+...) So, setting x->0 would leave you with just e^3. I'm still unsure of where exactly that e^3 pops up, but it may happen with the derivative (a series or something).

Answer 2

To do this problem you have to know his Therm

lim (1+x)^1/x = e
x->0

You're problem is
lim (1+3x)^1/x = e
x->0
Playing with this
let h=3x
Change everything to h, even the limit
as x->0 implies(from h=3x) that h->0
Substitute h for x and rewrite the equation in terms of h
lim (1+h)^3/h
h->0
or
lim [(1+h)^1/h]^3
h->0
since ^3 is a constant function, you can bring the limit inside
[lim (1+h)^1/h]^3
h->0
Now we know that inside is equal to e by the above definition, so the above reduces to
[e]^3 or e^3

Answer 3

Is 1/x an exponent...

I think the answer you're looking for is infinity.

EDIT: Well, infinity is the right-hand limit, I think 0 is the LH limit...

Answer 4

put the bong down and look at what you just typed...its so obvious, wow...the future of the united states

Answer 5

m