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anonymous
 5 years ago
By method od deltaepsilon, Prove that lim x>1(2x^23x) = 5
anonymous
 5 years ago
By method od deltaepsilon, Prove that lim x>1(2x^23x) = 5

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Is that suppose to be 2x^2+3x? because otherwise the limit is not 5!

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Analysis: Given \(\epsilon >0\) we would like to find \(\delta >0\) such that for all \(0<x1<\delta\) we have \(2x^2+3x5<\epsilon\) Notice that \(\begin{align*} 2x^2+3x5<\epsilon \iff 2x+5x1<\epsilon \end{align*}\) In order to be able to say something about x1 we need to bound the 2x+5 expression. If \(x1<1\) then by triangle inequality \(x1\leq x1<1\). Hence \(x<2\). It follows that \(2x+5\leq 2x+5<9\). So now in order to make \(2x+5x1<\epsilon\) it is enough by making \(9x1<\epsilon\), i.e. \(x1<\epsilon/9\). So which $\delta$ we need to choose to make \(x1<1\) and \(x1<\epsilon/9\) 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 $x1<\delta$ then $x1<1$. It follows that $x1<1$. Hence $x<2$. If follows that \(2x+5\leq 2x+5<2(2)+5=9\). Now \[2x^2+3x5=2x+5x1< 9x1<9(\epsilon/9)=\epsilon\]
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