anonymous
  • anonymous
4. Use the ǫ, δ-definition of limits to show that lim x!a f(x) = L if and only if lim x!a− f(x) = lim x!a+ f(x) = L.
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Please help me. See screenshot
1 Attachment
zzr0ck3r
  • zzr0ck3r
assume\[\lim_{x\rightarrow a}f(x)=L\]then by definition \[\forall\epsilon>0,\space \exists\space \delta >0,st\\a-\delta
zzr0ck3r
  • zzr0ck3r
so the converse is pretty much the same thing, just write out the definitions and the result follows.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Thanks! Could u help me look at this question? It makes me stunned.
1 Attachment
zzr0ck3r
  • zzr0ck3r
Give me a bit, I need cook something fast, then I will come back. Maybe close this and open another just in case others can help while im afk
zzr0ck3r
  • zzr0ck3r
im not sure this is right...
zzr0ck3r
  • zzr0ck3r
\[by\space definition\\\forall\epsilon>0,\exists\space \delta_1>0\space s.t.\space|x-a|<\delta_1\space \implies|g(x)-c|<\epsilon\\and\\\forall\space M\in R,\exists\space\delta_0>0\space s.t.\space0<|x-a|<\delta_0\space \implies f(x)>M\\let\space \delta \space=min\{\delta_0,\delta_0\}\\\text{then we have}\\|x-a|<\delta \implies|g(x)-c|<\epsilon \text{ and }f(x)>M\\so\\c-\epsilon

Looking for something else?

Not the answer you are looking for? Search for more explanations.