Find $\lim_{x\to 2}(x^2-3)$ using the definition of limit; also find $\delta$ for given $\epsilon$

Question

Find the limit $L$ using the definition of limit. Then find $\delta>0$ such that $|f(x)-L|<0.01$ whenever $0<|x-c|<\delta$. $$\lim_{x\to 2}(x^2-3)$$

I have gotten as far as $|x-2|=0.01/|x+2|$, and I need help with further steps, I do not know what more there is to do to find delta.

Source image

Answer 1

$$\lim_{x \rightarrow 2} (x^2-3)=2^2-3=4-3=1$$

The limit $L$ is equal to $1$.

We have to find a $\delta>0$ such that $|f(x)-1|<0.01$ whenever $0<|x-2|<\delta$.

EDIT: $$|x-2|<\delta \Rightarrow -\delta<x-2<\delta \Rightarrow 2-\delta<x<2+\delta$$

$$|f(x)-1|<0.01 \Rightarrow |x^2-3-1|<0.01 \Rightarrow |x^2-4|<0.01 \Rightarrow -0.01<x^2-4<0.01 \\ \Rightarrow 3.99<x^2<4.01 \Rightarrow \sqrt{3.99}<x<\sqrt{4.01}$$

Therefore, we could set $2+\delta=\sqrt{4.01} \Rightarrow \delta \approx 0.002$

Answer 2

If you choose $$\delta:=\min\{1,\frac\varepsilon5\}$$ then for $|x-2|<\delta$ you have $|x-2|<1$, which implies $1<x<3$ and $|x+2|<5$.

So you get $$|x^2-2|=|x+2|\cdot|x-2|<\frac\varepsilon5\cdot 5 = \varepsilon.$$