Basic question about limit

Question

I am new to limit. I have a function $f(x)=x^2$ for $x \neq 2$ and $f(2)=1$. I want to prove that $\lim_{x \to 2} f(x) = 4$. I am unsure about my work:

We want to show $\forall \epsilon > 0, \exists \delta >0$ such that $0<|x-2|<\delta \implies |f(x)-4|<\epsilon$.

$|f(x)-4| < \epsilon$

$|x^2-4| < \epsilon$

$-\epsilon < x^2 -4 < \epsilon$

$4-\epsilon < x^2 < 4+\epsilon$

$\sqrt{4-\epsilon} < x < \sqrt{4+\epsilon}$ for $\epsilon \leq 4$

$\sqrt{4-\epsilon} -2 < x-2 < \sqrt{4+\epsilon} -2$

We can choose $\delta = min( 2-\sqrt{4-\epsilon}, \sqrt{4+\epsilon}-2)$ for $\epsilon \leq 4$. For $\epsilon > 4$, we can choose a sufficiently small value of $\delta$ (say $\delta=0.0001$ )

Edit 1: Removed $|\sqrt{4-\epsilon}-2|< |x-2| < |\sqrt{4+\epsilon}-2|$

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Is it correct? Are there any thing to improve?

Answer 1

much simpler way to prove is following.first consider $\delta_1 = 1$ so $0<|x-2|<1$ implies $4<x+2<5$

$$|x^2-4| = |x+2|. |x-2|< 5.|x-2| < \epsilon \implies |x-2| < \frac{\epsilon}{5}$$

call this $\delta_2 = \epsilon/5.$ So we have that $\delta = \min\{1,\frac{\epsilon}{5}\}$

Answer 2

You are close. But what happens when you take $\epsilon = 0.1$ for instance? Also, why should $$|\sqrt{4-\epsilon}-2|< |x-2| < |\sqrt{4+\epsilon}-2|$$ imply $$\sqrt{4-\epsilon} -2 < x-2 < \sqrt{4+\epsilon} -2?$$

Edit. You may pick $\delta := \min{\{\sqrt{4 + \varepsilon} - 2, \sqrt{8}-2\}}$. To prove this $\delta$ works, you may need the inequality $$\sqrt{4-\varepsilon} < -\sqrt{4+\varepsilon}+4$$ for $0 < \varepsilon \leq 4.$

This inequality is also the reason why your choice of $\delta$ works for $\varepsilon\leq 4$; furthermore, it shows one of the terms in your $\min$ is not required as it is always bigger than the other.