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

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The discussion focuses on proving the inequality $2 - \epsilon < y < 2 + \epsilon$ for the function $y = 2 + \frac{1}{x^2}$, given that $x > \frac{1}{(\epsilon)^{1/2}}$. The first part of the proof establishes that if $x$ meets this condition, then $y < 2 + \epsilon$ holds true. Participants seek guidance on proving the lower bound, $2 - \epsilon < y$. It is noted that for any $x$, $y$ is always greater than 2, thus satisfying the condition $y > 2 > 2 - \epsilon$. The conversation emphasizes the need for a rigorous approach to complete the proof.
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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)
 
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Rido12 said:
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

$\chi$ $\sigma$
 
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