- #1
Miike012
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Prove: If f approaches l near a and f approaches m near a, then l = m.
...Im skipping to the end of the proof...
" to comlete the proof a particular ε>0 has to be choses for which the two conditions
|f(x) - l|< ε and |f(x) - m|< ε cannot both hold if l=/=m."
if l=/=m so that |l - m|> 0 , we can chose ε to be |l - m|/2... How did they decide this |l - m|/2??
It follows that there is a δ>0 such that for all x,
If 0<|x-a|<δ, then |f(x) - l|< |l - m|/2
and |f(x) - m|< |l - m|/2.
|l - m| = | l - f(x) + f(x) - m|≤ |l - f(x)| + | f(x) - m| < |l - m|/2 + |l - m|/2 = |l - m|, a contradiction.
can some one explain to me the contradiction?
...Im skipping to the end of the proof...
" to comlete the proof a particular ε>0 has to be choses for which the two conditions
|f(x) - l|< ε and |f(x) - m|< ε cannot both hold if l=/=m."
if l=/=m so that |l - m|> 0 , we can chose ε to be |l - m|/2... How did they decide this |l - m|/2??
It follows that there is a δ>0 such that for all x,
If 0<|x-a|<δ, then |f(x) - l|< |l - m|/2
and |f(x) - m|< |l - m|/2.
|l - m| = | l - f(x) + f(x) - m|≤ |l - f(x)| + | f(x) - m| < |l - m|/2 + |l - m|/2 = |l - m|, a contradiction.
can some one explain to me the contradiction?