How does the limit [1 + (1/x)]^x as x -> ± ∞ equal "e"? <-- Euler's number..
Because it gets closer and closer to 2.718.
\[(1+(1/x))^x\]
actually, it is [1 + 1/x]^x
yea amistre is right, so how..
How can I convince myself that this is true through algebra or calculus?
Calculus. :)
Hang on a second: how can you take the limit of 1/x inside the bracket, when the whole thing is being raised to the power of "x"?
it is 3 more or less by definition
?
i meant e of course
OK I have my piece of paper, pencil and calculator out right now. Tell me how I should proceed with convincing myself that this is true.
\[e=\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n\] is one definition
it depends on what your definition of "e" is. it it is the one i wrote, then there is nothing to prove.
still requires proof that the limit actually converges
click show steps
really if you want to make sense out of it you need to say what e is. the wolfram steps are silly in this case
lol
but zarkon is right. what you need to show is that it converges to something. it does, and we call that something "e" so to say it converges to e is a tautology, unless you have some other definition of e
you also said an algebraic way you can always see that the limit is getting close to e by taking large and larger values of x and plugging it in
argh you can show that it gets closer and closer to some number. if you want, you can call that number e
let's say \[e=\sum_{n=0}^\infty\frac{1}{n!}\] ;)
if i recall proof is not trivial.. you have to show it is increasing and bounded above. i think the bounded above part is easy though
very easy to show convergence with my def
how about \[\int_0^e\frac{dt}{t}=1\] as a definition?
that one is nice too
what gets my goat is calc problems like this where you are supposed to take the log, use l'hopital, and then exponentiate and say "oh look, i got e!" going around in circles
Join our real-time social learning platform and learn together with your friends!