Find $\lim_{x\to 2}(x^2-3)$ using the definition of limit; also find $\delta$ for given $\epsilon$ Find the limit $L$  using the definition of limit.  Then find $\delta&gt;0$ such that $|f(x)-L|&lt;0.01$ whenever $0&lt;|x-c|&lt;\delta$.
  $$\lim_{x\to 2}(x^2-3)$$


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.

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

So you get
$$|x^2-2|=|x+2|\cdot|x-2|&lt;\frac\varepsilon5\cdot 5 = \varepsilon.$$
$$\lim_{x \rightarrow 2} (x^2-3)=2^2-3=4-3=1$$

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

We have to find a $\delta&gt;0$ such that $|f(x)-1|&lt;0.01$ whenever $0&lt;|x-2|&lt;\delta$.

EDIT: 
$$|x-2|&lt;\delta \Rightarrow -\delta&lt;x-2&lt;\delta \Rightarrow 2-\delta&lt;x&lt;2+\delta$$

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

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