Friday, 26 April 2013

real analysis - Is there a way to show that ex=limntoinftyleft(1+fracxnright)n?



I know that e:=limn(1+1n)n,


by definition. Knowing that, I proved successively that ek=limn(1+kn)n,

when kN, kZ and kQ. Now, I was wondering : how can I extend this result over R ? I tried to prove that fn(x):=(1+xn)n converge uniformly on R but unfortunately it failed (I'm not sure that it's even true). Any idea ?







My idea was to define the function xex as ex={exxQlimneknif knx and (kn)Q.


But to conclude that ex=limn(1+xn)n,

I need to prove that fn(x)=(1+xn)n converge uniformly on a neighborhood of x, but I can't do it. I set gn(x)=fn(x)ex,

but I can't find the maximum on a compact that contain x, and thus can't conclude.


Answer



We can use that exists pn,qnQ such that pn,qnx and pnxqn, therefore



(1+pnn)n(1+xn)n(1+qnn)n




and



(1+pnn)n=[(1+pnn)npn]pnex



(1+qnn)n=[(1+qnn)nqn]qnex



indeed for npn(m,m+1) with mN we have



(1+1m+1)m(1+pnn)npn(1+1m)m+1




and therefore (1+pnn)npne.


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