- #1
transgalactic
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the question:
f(x) continues on (-\infty,a]
and suppose that the border \lim_{x->-\infty}f(x) exists and finite.
prove that f(x) is bounded on (-\infty,a] and/or that exists
x_0\epsilon(-\infty,a]=\lim _{x->-\infty }f(x)
so
\sup_{x\epsilon(-\infty,a]} f(x)
in other words prove that f(x) gets its highest value on (-\infty,a]
and the supremum is the maximum
the non understood part:
suppose
\lim_{x->-\infty}f(x)=m_0
suppose m_0<a
and we check on the interval of [m_0,a] where [m_0,a]\subseteq (-\infty,a]
they prove by a counter example that:
"suppose the function is not bounded from the top then \forall n\epsilon N and
m_0\leq x_n\leq a"
i can't understand it.if a function is bounded by some epsilon then we take N for which after this N (n>N) f(x)<epsilon
if its not bounded from the top then
f(x) is bigger then epsilon for the whole interval
this is not what they writee up there
what are they writing there??
f(x) continues on (-\infty,a]
and suppose that the border \lim_{x->-\infty}f(x) exists and finite.
prove that f(x) is bounded on (-\infty,a] and/or that exists
x_0\epsilon(-\infty,a]=\lim _{x->-\infty }f(x)
so
\sup_{x\epsilon(-\infty,a]} f(x)
in other words prove that f(x) gets its highest value on (-\infty,a]
and the supremum is the maximum
the non understood part:
suppose
\lim_{x->-\infty}f(x)=m_0
suppose m_0<a
and we check on the interval of [m_0,a] where [m_0,a]\subseteq (-\infty,a]
they prove by a counter example that:
"suppose the function is not bounded from the top then \forall n\epsilon N and
m_0\leq x_n\leq a"
i can't understand it.if a function is bounded by some epsilon then we take N for which after this N (n>N) f(x)<epsilon
if its not bounded from the top then
f(x) is bigger then epsilon for the whole interval
this is not what they writee up there
what are they writing there??
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