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.
Best Answer
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.