<s>[INST] Prove that $\lim_{x\rightarrow c}x^2=c^2$ What is the common proof for $\lim_{x\rightarrow c}x^2=c^2$.
I tried like this
Let $\varepsilon>0$, want to find a $\delta>0$ such that $\forall x\in\mathbb{R},0<|x-c|<\delta\Rightarrow |x^2-c^2|<\varepsilon$.
Therefore $$-\varepsilon<x^2-c^2<\varepsilon$$
$$c^2-\varepsilon<x^2<\varepsilon+c^2$$
$$-c-\sqrt{c^2+\varepsilon}<x-c<-c+\sqrt{c^2+\varepsilon}$$
Choose $\delta=-c+\sqrt{c^2+\varepsilon}$which proves the problem
Is this proof is correct ? If there is any mistake in it give a correct proof please .thanks
 [/INST] $$lim_{x \to c} x^2 = c^2$$
Let $\epsilon>0$ , we know that
$$x^2-c^2=(x-c)(x+c)$$
$$|x^2-c^2|=|x-c||x+c|$$
If
$$|x-c|<2 \implies -2<|x|-|c|<2 (by\ \ triangular \ \ property)$$
$$|c|-2<|x|<2+|c|$$
By triangular property, we have
$$|x+c|\le |x|+|c|<2+|c|+|c|$$
$$|x+c|<2+2|c|$$
So,
$$|x^2-c^2|<2.(2+2|c|)$$
$$|x^2-c^2|<(4+4|c|)$$
$$|x^2-c^2|<(2+2|c|)(|x-c|)$$
If $$|x-c|<\frac{\epsilon}{2+2|c|}$$
Then
$$|x^2-c^2|<\epsilon$$
Define
$$\delta (\epsilon) = inf \{2, \frac{\epsilon}{2+2|c|}\}$$
Then if $|x-c|< \delta \implies |x^2-c^2|<\epsilon$
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