Formal proof of limit By using $\epsilon-\delta$ definition, prove that $$\lim_{x\rightarrow1}\frac{3}{x^2}=3$$

Here is what I have tried.

$$|\frac{3}{x^2}-3|=3|\frac{1}{x^2}-1|=3|\frac{1-x^2}{x^2}|=3\frac{|x-1||x+1|}{x^2}$$

So I have problem with how to elimate the denominator.
Given $\epsilon &gt; 0$, choose $\delta = \text{min}\left(\frac{1}{2},\frac{\epsilon}{30}\right) \to |1|-|x| &lt; |1-x|=|x-1| &lt; \delta &lt; \dfrac{1}{2} \to |x| &gt; \dfrac{1}{2}$, and $|x+1| = |x-1+2| &lt; |x-1|+|2| &lt; \delta + 2 &lt; \dfrac{1}{2} + 2 = \dfrac{5}{2} \to \left|\dfrac{3}{x^2} - 3\right| &lt; 3\cdot |x-1|\cdot \dfrac{5}{2}\cdot \dfrac{1}{\left(\frac{1}{2}\right)^2}=30|x-1|&lt; \epsilon$
Since $x$ is close to $1$ then $1/x^2 &lt; 4$ and $|x+1|&lt;3$.