$$|x^2-1|< 1/n$$ for a give n is a natural number.
Here is my own solution, please check it for me: $$|x^2 -1|=|x+1||x-1| \leq (|x+1|)|x-1|.$$ $$ |x|=|x-1|+1.$$ we chose $$\delta_1=1$$ Then if $$ |x-1|\lt\delta_1=1 $$ We have $$|x|\leq|x-1|+1<1+1=2. $$$$|x^2 -1|=|x+1||x-1| \leq (|x+1|)|x-1|\leq (2+1)|x-1|$$. We choose $$\delta=\min\left(1,\frac1{3n}\right) = \frac1{3n}$$ Then $$|x^2-1|\leq3|x-1|<3\cdot\left(\frac1{3n}\right)=\frac1n$$ Thus, if$$|x-1|<\frac1{3n}$$ Then $$|x^2-1|<\frac1n$$
Your answer is good.
Along the same lines, for $n\ge1$, the triangle inequality says $$ \underbrace{|x-1|\le\frac2{5n}}_{\substack{\text{weaker than}\\|x-1|\le\frac1{3n}}}\implies|x+1|\le2+\frac2{5n}\le\frac{12}5\implies\underbrace{|x^2-1|\le\frac{24}{25n}}_{\substack{\text{stronger than}\\|x^2-1|\le\frac1n}} $$
Great solution. I don't think I would have done better myself. In the first line, you wrote $$\ldots |x + 1||x - 1| \leq (|x+1|)|x-1|,$$
but you could have written $$|x+1||x-1| \leq (|x|+1)|x-1|$$
at once. This is not so important, since you got it right in the end. And you can use \leq to produce $\leq$ instead of $<=$. (And you might find helpful my answer here, it is a bit of a general strategy for dealing with $\epsilon -\delta$ proofs with polynomials.)
