Limit problem involving proofs

Question

Given that

$$\lim_{x\to 0}\; (3x + 1)(3x - 1)x^2 + 0.01=0.01$$

Prove that there exists an open interval $(a, b)$ containing $0$ such that $(3x + 1)(3x - 1)x^2 + 0.01 > 0$ for all $x$ not equal to zero in $(a, b)$.

Answer 1

Suppose that $f: \mathbb{R} \to \mathbb{R}$. Recall that by the definition of limit, we have that if $\lim\limits_{x \to a} f(x) = L$, then for every $\varepsilon >0$, there exists a $\delta>0$ such that if $0 < |x-a| < \delta$ it follows that $|f(x) - L| < \varepsilon$.

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With this problem, we are given that $\lim\limits_{x \to a} (3x+1)(3x-1)x^2 + .01 = .01$.

From the definition, let's take $\varepsilon = .01$. Then we have that there exists a $\delta >0$ such that for all $x$ satisfying $0 < |x| < \delta$ we have $|((3x+1)(3x-1)x^2 + .01) - .01| < .01$.

That is, for $x \in (0-\delta, 0 + \delta)$, we have

$$ -.01 < (3x+1)(3x-1)x^2 < .01. $$ And thus in that interval, $0 < (3x+1)(3x-1)x^2 + .01 < .02$.