Epsilon delta proof for continuity

Question

I am trying to do some rough work for an epsilon delta proof, that proves $f(x)=x^4+2x^2$ is continuous at $x=-1$. Working backwards I am trying to find a delta in terms of epsilon to start my proof.

So $$|f(x)-f(a)|= |x^4+2x^2-3| \lt \epsilon$$ Then $$|x^4+2x^2-3| = |x+1||x-1||x^2+3| = \delta |x-1||x^2+3| \lt \epsilon$$ Also we have $|x+1|\lt \delta$ so $|x-1| \lt \delta-2$:

Therefore: $$\delta(\delta-2)|x^2+3| \lt \epsilon$$

However now I am struggling to find a delta expression for $x^2+3$. Is my working so far right? If so could someone show me how to go further!

Answer 1

Since you have to investigate the function near $-1$, we can assume that $-2<x<0.$

Then we have $|x-1|<3$ and $|x|<2$, hence $x^2+3 <7.$

This gives:

$$|x^4+2x^2-3| = |x+1||x-1||x^2+3| \le 3 \cdot 7 \cdot |x+1|=21|x+1|.$$

Can you take it from here ?

Answer 2

Note that\begin{align}\lvert x^2+2x^2-3\rvert&=\lvert x^4-1+2x^2-2\rvert\\&\leqslant\lvert x^4-1\rvert+\lvert2x^2-2\rvert\\&=\lvert x-1\rvert.\rvert x^3+x^2+x+1\rvert+2\lvert x-1\rvert.\lvert x+1\rvert.\end{align}So, take $\delta=\min\left\{1,\frac\varepsilon8\right\}$. Then, if $\lvert x-1\rvert<\delta$, since $\delta\leqslant1$ we have$$\lvert x^3+x^2+x+1\rvert\leqslant4\text{ and }\lvert x+1\rvert\leqslant2.$$So$$\lvert x^2+2x^2-3\rvert\leqslant4\lvert x-1\rvert+4\lvert x-1\rvert=8\lvert x-1\rvert.$$And now, since $\delta\leqslant\frac\varepsilon8$,$$\lvert x-1\rvert<\delta\implies8\lvert x-1\rvert<\varepsilon.$$