Prove: If f approaches l near a and f approaches m near a, then l = m.(adsbygoogle = window.adsbygoogle || []).push({});

....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?

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# Homework Help: Limit proof by contradiction

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