I know that e:=limn→∞(1+1n)n,
by definition. Knowing that, I proved successively that ek=limn→∞(1+kn)n,
when k∈N, k∈Z and k∈Q. 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 x⟼ex as ex={exx∈Qlimn→∞eknif kn→x 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,qn∈Q such that pn,qn→x and pn≤x≤qn, therefore
(1+pnn)n≤(1+xn)n≤(1+qnn)n
and
(1+pnn)n=[(1+pnn)npn]pn→ex
(1+qnn)n=[(1+qnn)nqn]qn→ex
indeed for npn∈(m,m+1) with m∈N we have
(1+1m+1)m≤(1+pnn)npn≤(1+1m)m+1
and therefore (1+pnn)npn→e.
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