Prove $\epsilon - \delta$ style that $\lim\limits_{x \rightarrow 2}x^2 \neq 6$ via contradiction

Question

Question: Prove $\epsilon - \delta$ style that $\lim\limits_{x \rightarrow 2}x^2 \neq 6$ via contradiction

So my initial idea is to assume $\lim\limits_{x \rightarrow 2} x^2 = 6$. Then for all $\epsilon > 0$ $\exists$ $\delta > 0$ such that $|x^2-6| < \epsilon \rightarrow0 < |x-2| < \delta$

However, I am not sure how to show a contradictio without "plugging it in".... could someone show me?

Answer 1

Let $\varepsilon = 0.25 > 0 $

We have that for all $\delta > 0$, if we take $\alpha = \text{min}\{0.1,\frac{\delta}{2} \}$, then we have that $2\alpha + \alpha^{2} \leq 0.2 + 0.01 = 0.21$

If we take $x = 2 + \alpha$, we have that $|2+ \alpha - 2| = \alpha < \delta$, but $|(2+ \alpha)^{2} - 6| = 6 - 4 - 2\alpha - \alpha^{2} \geq 2 - 0.21 > 1 > \varepsilon$. Then is a contradiction of the limit definition