# Proving $\lim_{x\to 3} (x^2-5x+1)=-5$ by the $\epsilon -\delta$ definition of a limit.

Prove that $\displaystyle\lim_{x\to 3} (x^2-5x+1)=-5$ by the $\epsilon -\delta$ definition of a limit.

What I've done so far:

$\forall \epsilon >0 \exists \delta\ni 0<|x-3|<\delta\Rightarrow 0<|x^2-5x+1-5|<\epsilon\\\Rightarrow 0<|x^2-5x+6|<\epsilon\Rightarrow 0<|x-2||x-3|<\epsilon$ $|x-2|=|x-3+1|\Rightarrow |x-2|\leq |x-3|+1$ (by the Triangle Inequality.)

So $|x-2||x-3|\leq|x-3|(|x-3|+1)<\epsilon\Rightarrow |x-2||x-3|\leq \delta (\delta +1)=\delta ^2 +\delta <\epsilon\\\Rightarrow\delta ^2+\delta-\epsilon <0$

By the quadratic formula, $\delta =\dfrac{-1+\sqrt{1+4\epsilon}}{2}$.

I'm not sure how to proceed from here. Is the value of $\delta$ obtained from the quadratic formula correct? If not, how do I continue? Thanks.

• This is unnecessary overcomplication. You can let $\delta < 1$ and so $\delta (\delta + 1) < 2\delta$. Hence, you can simply choose $\delta = \text{min}(\frac{\varepsilon}{2},1)$ and leave it at that. I presume the arithmetic is correct up to that point. – Christopher K Nov 22 '13 at 4:24
• @ChrisK, I approve emphatically :) – Mr.Fry Nov 22 '13 at 4:28
• @ChrisK, could you explain a little more? Why would $\delta=\min\left(\dfrac{\epsilon}{2},1\right)$? – Sujaan Kunalan Nov 22 '13 at 4:32
• Well, we let $\delta<1$ as well as $\delta (\delta + 1) < 2\delta < \varepsilon$. So, $\delta < \frac{\varepsilon}{2}$ and hence all we need to choose is $\delta = min(\frac{\varepsilon}{2},1)$. – Christopher K Nov 22 '13 at 4:43
• – Mhenni Benghorbal Nov 22 '13 at 4:50

Hint: You want to make $|f(x)-f(3)|$ look like $|x-3|$ then make an assumption about $|x-3|$ say $|x-3|< \delta \leq 1$ which will give you a bound for x. Comment if you need help.
• How do you know $\delta\leq 1$? – Sujaan Kunalan Nov 22 '13 at 4:39
• You can assume an upper bound for $\delta$ you could of chosen $1,2,3,..,n$ just we want to control this x. We don't want x to blow up. – Mr.Fry Nov 22 '13 at 4:40
• BY definition we say is we can get $|x-a|<\delta$ then we have that $|f(x)-f(a)|<\epsilon$ – Mr.Fry Nov 22 '13 at 4:43