Prove $\lim_{x\to\sqrt3}\frac1{x^2}=\frac13$ using the epsilon–delta definition.
Here is my work. Is it correct? $$\left|\frac1{x^2}-\frac13\right|<\varepsilon\quad\quad|x-\sqrt3|<\delta$$ $$\left|\frac{(\sqrt3-x)(\sqrt3+x)}{3x^2}\right|<\varepsilon$$ $$\delta=1:|x-\sqrt3|<1$$ $$\sqrt3-1<x<\sqrt3+1$$ $$3(\sqrt3-1)^2<3x^2<3(\sqrt3+1)^2$$ $$\text{also: }2\sqrt3-1<x+\sqrt3<2\sqrt3+1$$ $$\therefore\left|\frac{(\sqrt3-x)(2\sqrt3+1)}{3(\sqrt3-1)^2}\right|<\varepsilon$$ $$|\sqrt3-x|<\frac{\varepsilon(3(\sqrt3-1)^2)}{2\sqrt3+1}$$ $$\therefore\frac{\varepsilon(3(\sqrt3-1)^2)}{2\sqrt3+1}=\delta$$ $$\left|\frac1{x^2}-\frac13\right|<\varepsilon$$