anonymous
  • anonymous
can someone tell me how to prove for infinite limit with epsilon and delta or N and M?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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abb0t
  • abb0t
Yes, if you gave us an example, we could.
abb0t
  • abb0t
What's the function?
anonymous
  • anonymous
l\[\lim_{x \rightarrow \infty}3x^{2}-x-2/5x^{2}+4x+1\]

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ganeshie8
  • ganeshie8
\[\lim_{x \rightarrow \infty}~\dfrac{3x^{2}-x-2}{5x^{2}+4x+1}\] like that ?
anonymous
  • anonymous
Yes, like that
ganeshie8
  • ganeshie8
so what does it evaluate to ?
ganeshie8
  • ganeshie8
you need to know the limit before starting to prove it using epsilon-delta
anonymous
  • anonymous
i know the limit it's 3/5
anonymous
  • anonymous
but the prove is quite confusing.
ganeshie8
  • ganeshie8
its easy once you know how to play the game
anonymous
  • anonymous
i guess... so :)
ganeshie8
  • ganeshie8
Let \(f(x) =\dfrac{3x^{2}-x-2}{5x^{2}+4x+1} \) given any positive number \(\epsilon\), you have to show that there exists some \(x\) value, after which the difference between \(f(x)\) and \(\dfrac{3}{5}\) is less than \(\epsilon\).
ganeshie8
  • ganeshie8
Given any positive number \(\epsilon\), prove that there exists some \(x\) value beyond which below is true : \(\left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\lt \epsilon\)
ganeshie8
  • ganeshie8
simplify and try expressing \(x\) in terms of \(\epsilon\)
ganeshie8
  • ganeshie8
@kal1921
anonymous
  • anonymous
simplify? sorry where is delta?
zzr0ck3r
  • zzr0ck3r
Simplify so that you can find the bound on delta. You have to work backwards
ganeshie8
  • ganeshie8
yeah what we're doing is just scratch work to express \(x\) in terms of \(\epsilon\)
anonymous
  • anonymous
\[-\epsilon< equation<\epsilon\] i get this...what is next?
ganeshie8
  • ganeshie8
for \(x\gt 1\), we have : \( \left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\\~\\ =\left|\dfrac{17x+13}{5(5x^2+4x+1)}\right|\\~\\ \lt\left|\dfrac{17x+13}{25x^2}\right|\\~\\ \lt\left|\dfrac{17x+17x}{25x^2}\right|\\~\\ =\left|\dfrac{34}{25x}\right|\\~\\ \) \(\dfrac{34}{25x} \lt \epsilon \implies x \gt \dfrac{34}{25\epsilon}\) therefore \(x = \dfrac{34}{25\epsilon}\) works, now you can start the proof.
anonymous
  • anonymous
sorry, how can i relate x and delta... if it x wasn't approaching to infinity i would say /x-a/>delta
ganeshie8
  • ganeshie8
The definition for limits at infinity is slightly different : \(\lim\limits_{x\to\infty}f(x)=L\) means that for "every" \(\epsilon\gt 0\), there exists some \(x\) value, \(N\), such that \(|f(x)-L|\lt\epsilon\) for all \(x\gt N\).
ganeshie8
  • ganeshie8
In present problem, we have cooked up that \(N\) to be \(\dfrac{34}{25\epsilon}\)
ganeshie8
  • ganeshie8
you choose "any" positive \(\epsilon\), the value \(|f(x)-\frac{3}{5}|\) will always be less than that chosen \(\epsilon\) whenever \(x\) is greater than \(\dfrac{34}{25\epsilon}\)
ganeshie8
  • ganeshie8
In other words, we can make \(f(x)\) as close to \(\frac{3}{5}\) as we want by making \(x\) sufficiently large.
ganeshie8
  • ganeshie8
I think this is same as the N and M definition that you were refering to in main question

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