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

1. watchmath

Is that suppose to be 2x^2+3x? because otherwise the limit is not 5!

2. anonymous

it is

3. anonymous

no wait one second

4. anonymous

yes its 2x^2+3x

5. anonymous

hi

6. 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 $|2x^2+3x-5|=|2x+5||x-1|< 9|x-1|<9(\epsilon/9)=\epsilon$