Question

lim x-->0+ [1+(1/n)]^n

Answer 1

This is a definition of \(e\), isn't it?

Answer 2

\[ \lim_{x \rightarrow 0}(1+\frac{ 1 }{ n })^n\]

Answer 3

You might wanna try using a logarithm...

Answer 4

as in ln(1+1/n)^n = n ln (1+1/n)

Answer 5

\[
1+\frac{1}{n}=\frac{n}{n}+\frac{1}{n} = \frac{n+1}{n}
\]

Answer 6

is the n in front of the ln correct?

Answer 7

Yes, but you also have to use the property for division inside logarithms.

Answer 8

yes i see that now

Answer 9

Then it is continuous. Remember the whole thing should be a power of \(e\).

Answer 10

or maybe no continuous but rather... what's the term...

Answer 11

but i can't have a n in the denominator with n approachhing zero

Answer 12

One way I think about that part is to assume limit evaluates to some value L, then I take log of both sides.

Answer 13

If you use the division property for logarithms, it is no longer divided.

Answer 14

so its n[ln(n+1)-ln(n)]?

Answer 15

yes, no?

Answer 16

Yes.

Answer 17

well what do we do with that n on the outside? it's negating the entire equation

Answer 18

Nothing...

Answer 19

that would give us ln y = 0

Answer 20

I'm much more concerned with \(\ln(n), n\to 0\) because

Answer 21

It doesn't exist going from the left.

Answer 22

i don't understand, is that what that plus is for?

Answer 23

ok i'm at \[lny=\lim_{x \rightarrow 0+} n]\ln(n+1)-\ln(n)]\]

Answer 24

am i anywhere near the right track?

Answer 25

actually there is no limit... because if x===> 0 in an expression without x's you don't need to do anything

just an observation...

Answer 26

i meant n->0+

Answer 27

now can you help?