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anonymous

  • 5 years ago

By method od delta-epsilon, Prove that lim x->1(2x^2-3x) = 5

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  1. watchmath
    • 5 years ago
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    Is that suppose to be 2x^2+3x? because otherwise the limit is not 5!

  2. anonymous
    • 5 years ago
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    it is

  3. anonymous
    • 5 years ago
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    no wait one second

  4. anonymous
    • 5 years ago
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    yes its 2x^2+3x

  5. anonymous
    • 5 years ago
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    hi

  6. watchmath
    • 5 years ago
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    Analysis: Given \(\epsilon >0\) we would like to find \(\delta >0\) such that for all \(0<|x-1|<\delta\) we have \(|2x^2+3x-5|<\epsilon\) Notice that \(\begin{align*} |2x^2+3x-5|<\epsilon \iff |2x+5||x-1|<\epsilon \end{align*}\) In order to be able to say something about |x-1| we need to bound the |2x+5| expression. If \(|x-1|<1\) then by triangle inequality \(|x|-1\leq |x-1|<1\). Hence \(|x|<2\). It follows that \(|2x+5|\leq 2|x|+5<9\). So now in order to make \(|2x+5||x-1|<\epsilon\) it is enough by making \(9|x-1|<\epsilon\), i.e. \(|x-1|<\epsilon/9\). So which $\delta$ we need to choose to make \(|x-1|<1\) and \(|x-1|<\epsilon/9\) at the same time? Choose \(\delta:=\min \{1,\epsilon/9\}\). Here is the rigorous proof (the analysis above is to come up with the following proof) Given \(\epsilon >0\) choose \(\delta:=\{1,\epsilon/9\}\). Since $\delta <1$, if $|x-1|<\delta$ then $|x-1|<1$. It follows that $|x|-1<1$. Hence $|x|<2$. If follows that \(|2x+5|\leq 2|x|+5<2(2)+5=9\). Now \[|2x^2+3x-5|=|2x+5||x-1|< 9|x-1|<9(\epsilon/9)=\epsilon\]

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