anonymous
  • anonymous
Prove the statement using ɛ, ɓ definition of a limit. Lim (1+(1/3)x)=2 x->3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
definition in this case would be given any \(\epsilon>0\) there is a \(\delta\) which you write as a function of \(\epsilon\) that if \(|x-3|<\delta\) then \(|1+\frac{1}{3}x-2|<\epsilon\)
anonymous
  • anonymous
work backwards \[|\frac{1}{3}x-1|<\epsilon\]\[-\epsilon<\frac{1}{3}x-1<\epsilon\]\[1-\epsilon<\frac{1}{3}x<\epsilon+1\] \[3-3\epsilon
anonymous
  • anonymous
the last one is the same as \[|x-3|<3\epsilon\] to you can say "let \(\delta=3\epsilon\)" work the algebra backwards and you have your proof

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anonymous
  • anonymous
not really a coincident that your line with slope \(m\) requires a \(\delta\) of \(\frac{1}{m}\)
anonymous
  • anonymous
Is there any easy way to do/explain this cause I don't really understand the steps that you are doing.
zzr0ck3r
  • zzr0ck3r
well ok we need to show that if |x-3| < delta then |f(x)-L|
zzr0ck3r
  • zzr0ck3r
i dont think I made anything clearer but I gave it a shot
anonymous
  • anonymous
Just wondering where he got the \[|x-3| < \delta\]
zzr0ck3r
  • zzr0ck3r
\[|\frac{1}{3}x-1|<\epsilon\\|\frac{x-3}{3}|<\epsilon\\\frac{|x-3|}{|3|}<|x-3|<\epsilon\]
zzr0ck3r
  • zzr0ck3r
\[so\space given\space \epsilon\\let\space\delta=\epsilon\\if\space|x-3|<\delta\\then\\|1+\frac{1}{3}x-2|<\frac{|x-3|}{3}<|x-3| <\epsilon\]

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