anonymous
  • anonymous
see attached question Prove the following limits using only the ǫ, δ-definition.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
\[\forall \epsilon >0\;\exists \delta\; \forall x:\quad x>\delta\implies \left|\frac{x^2+2x}{x^2+1}-1\right| <\epsilon \]
anonymous
  • anonymous
\[ \left|\frac{x^2+2x}{x^2+1}-1\right| = \left|\frac{x^2+2x}{x^2+1}-\frac{x^2+1}{x^2+1}\right| = \left|\frac{2x-1}{x^2+1}\right| \]

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anonymous
  • anonymous
i got choose δ = 2/ϵ. do u get this answer
anonymous
  • anonymous
How did you get that answer?
anonymous
  • anonymous
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anonymous
  • anonymous
Doesn't really explain \(2/\epsilon\)
anonymous
  • anonymous
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anonymous
  • anonymous
is that right? what do u think ?
anonymous
  • anonymous
the lesser the denominator, the greater the fraction; the greater the numerator, the greater the fraction.
anonymous
  • anonymous
Wait, don't you want it to be smaller rather than bigger?
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
You don't want to over estimate their \(\epsilon\), you want to underestimate it.
anonymous
  • anonymous
If they give you \(\epsilon=1\) then it's not safe to think \(\epsilon = 2\) while it is safe to think \(\epsilon=1/2\)
anonymous
  • anonymous
okay. what is the answer then? Help me
anonymous
  • anonymous
However, I think you were on the right course. Clearly we have something similar to \(2/x\)
anonymous
  • anonymous
I didn't really over estimate ϵ actually. we treate ϵ is as a constant..... otherwise no point to write δ in terms of ϵ
anonymous
  • anonymous
The point is, it is not safe to have the fraction getting larger, you want it to be getting smaller.
anonymous
  • anonymous
That way it remains in the epsilon range.
anonymous
  • anonymous
the open statement is given ϵ >0, there exists δ >0 such that x >δ for all x, then the limit. we have treated ϵ as a constant in the beginning. we need to find δ
anonymous
  • anonymous
I know what you mean. I'll double check. if you have the answer. Please let me know .thanks
anonymous
  • anonymous
Let's say \(\delta = 1\implies x>1\):\[ \left|\frac{2x-1}{x^2+1}\right| > \left|\frac{2x-x}{x^2+x}\right|=\left|\frac{x}{x(x+1)}\right|=\left|\frac{1}{x+1}\right|>\left|\frac{1}{2x}\right| \]
anonymous
  • anonymous
\[ \frac{1}{2x} < \epsilon \]
anonymous
  • anonymous
So we could say: \[ \frac{1}{2\epsilon} < \delta \]When we know \(\delta >1\)
anonymous
  • anonymous
Then I guess the result is \[\delta = f(\epsilon) =\min\left\{1,\frac{1}{2\epsilon}\right\}\]
anonymous
  • anonymous
so u choose a random number. what about δ=100
anonymous
  • anonymous
What about it?
anonymous
  • anonymous
well. not really like that
anonymous
  • anonymous
I was wrong, I should have said \(\max\)
anonymous
  • anonymous
please explain this step to me
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anonymous
  • anonymous
It should be: \[\delta = f(\epsilon) =\max\left\{1,\frac{1}{2\epsilon}\right\}\]
anonymous
  • anonymous
yes, maybe u should say max.
anonymous
  • anonymous
\[ \frac{1}{2x}<\epsilon \implies \frac{1}{2}<\epsilon x\implies \frac{1}{2\epsilon}

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