Proving $2- \epsilon < y$ for Limit w/ Epsilon

[Dethrone]

Suppose we are given the function $y=2+\frac{1}{x^2}$. Prove that given $x>\frac{1}{(\epsilon)^{1/2}}$, where $\epsilon > 0$, then $2- \epsilon < y < 2 + \epsilon$.
So the first part is easy:
$$x>\frac{1}{(\epsilon)^{1/2}}$$
$$x^2>\frac{1}{\epsilon}$$
$$\frac{1}{x^2}<\epsilon$$
$$2+\frac{1}{x^2}<2+\epsilon$$

Now, any hints as to how to prove $2-\epsilon < y$? (Wondering)

 

[chisigma]

Suppose we are given the function $y=2+\frac{1}{x^2}$. Prove that given $x>\frac{1}{(\epsilon)^{1/2}}$, where $\epsilon > 0$, then $2- \epsilon < y < 2 + \epsilon$.
So the first part is easy:
$$x>\frac{1}{(\epsilon)^{1/2}}$$
$$x^2>\frac{1}{\epsilon}$$
$$\frac{1}{x^2}<\epsilon$$
$$2+\frac{1}{x^2}<2+\epsilon$$

Now, any hints as to how to prove $2-\epsilon < y$? (Wondering)

For any value of x and for any value of $\varepsilon > 0$ is ...

$\displaystyle y > 2 > 2 - \varepsilon\ (1)$

Kind regards

$\chi$ $\sigma$