Find $\delta > 0$ such that $|x-2| < \delta$ implies $|x^2+2x-18| < \frac{1}{4}$ Find $\delta > 0$ such that $|x-2|  < \delta$ implies $|x^2+2x-18| < \frac{1}{4}$.
I have been having trouble with this question from my Analysis class because $f(x)$ doesn't factor. Here is the first scratch work I did attempting to find a suitable $\delta$:
Let $\delta_1 =1$. Then $|x-2| < \delta \implies -1 < x-2<1 \implies 3 < x+2 < 5$. Hence $|x+2|<5$.
We want $|x^2+2x-18|< \frac{1}{4}$. This is the same as:
$$\frac{-1}{4} < x^2+2x-18 < \frac{1}{4}$$
$$-1 < 4x^2+8x-72 < 1$$
$$71 < 4x^2+8x < 73$$
$$71 < 4x(x+2) < 73$$
$$|4x(x+2)|=|4x||x+2| \leq |4x| \cdot5 < 73$$
$$|4x| < \frac{73}{5}$$
I got stuck here. In the other practice problems we did I always ended up with something like $|x-a|<...<...$ so I'm not sure what went wrong? I do not see how to factor $x^2+2x-18$ into something that utilizes $(x-2)$.

Let $f(x)=|x^2+2x-18|$. For all $\delta>0$, $|2-2|<\delta$ but $f(2)=10\geq\frac{1}{4}$. So no matter which value you pick for $\delta$ there is a $x$ ($2$) that wont satisfy $|x^2+2x-18|<\frac{1}{4}$.
