anonymous
  • anonymous
By method od delta-epsilon, Prove that lim x->1(2x^2-3x) = 5
Mathematics
  • 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!
watchmath
  • watchmath
Is that suppose to be 2x^2+3x? because otherwise the limit is not 5!
anonymous
  • anonymous
it is
anonymous
  • anonymous
no wait one second

Looking for something else?

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

More answers

anonymous
  • anonymous
yes its 2x^2+3x
anonymous
  • anonymous
hi
watchmath
  • watchmath
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 \[|2x^2+3x-5|=|2x+5||x-1|< 9|x-1|<9(\epsilon/9)=\epsilon\]

Looking for something else?

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