Question

Prove that the limit x tends to 1 (2x^4-6x^3+x^2+3)÷(x-1)=-8

Answer 1

First note that if $x\neq1$, you have

$\dfrac{2x^4-6x^3+x^2+3}{x-1}=2x^3-4x^2-3x-3$

Now, you're looking for $\delta>0$ such that for any $\varepsilon>0$, you have

$|x-1|$

Note that you can divide through the left side of the $\varepsilon$ inequality by $x-1$ once more:

$\dfrac{2x^3-4x^2-3x+5}{x-1}=2x^3-2x-5\implies 2x^3-4x^2-3x+5=(x-1)(2x^2-2x-5)$

So it follows that you need to find an appropriate $\delta$ that will guarantee

$|(x-1)(2x^2-2x-5)|=|x-1||2x^2-2-5|$

For the moment, let's fix $\delta=1$. Then by this assumption, we have

$|x-1|$

From this we get

$\implies0$
$\implies0$
$\implies0$
$\implies-5$
$\implies1$

where the upper bound is what we care about. With this assumption, we then get that

$|x-1||2x^2-2-5|$

which suggests that $\delta$ can be taken to be either the smaller of 1 or $\dfrac{\varepsilon}5$, or $\delta=\min\left\{\dfrac{\varepsilon}5,1\right\}$, to guarantee that the function gets arbitrarily close to -8.