Prove the limit of the below function with epsilon-delta definition

Question

How do I prove the limit of following function using the epsilon-delta definition 1

Answer 1

We need to prove for $\epsilon>0$ there is $\delta>0$ such that $|x-2|<\delta \Longrightarrow |x^2-4|<\epsilon$.

Let us start by $|x^2-4|<\epsilon \Longrightarrow |x-2|\cdot |x+2|<\epsilon \Longrightarrow |x-2|<\frac{\epsilon}{|x+2|}$.

The limit is only dependent near the point so that we can take $\delta = Min\{\epsilon/5,1\} \leq 1$. We need to show, it satisfies the condition $|x-2|<\delta \Longrightarrow |x^2-4|<\epsilon$.

If $\epsilon \geq 5$ then $|x-2|<\delta = 1 \Longrightarrow x\in (1,3) \Longrightarrow |x+2|<5 \Longrightarrow |x^2-4|= |x-2|\cdot |x+2| < 5 \leq \epsilon$.

If $\epsilon < 5$ then $|x-2|<\delta = \frac{\epsilon}{5} < 1 \Longrightarrow x\in (1,3) \Longrightarrow |x+2|<5 \Longrightarrow |x^2-4|= |x-2|\cdot |x+2| < 5 \delta = 5\frac{\epsilon}{5} =\epsilon$.