Certainly! Let's solve the inequality step-by-step:
The given inequality is:
[tex]\[ -\varepsilon < \frac{1}{x^2} - \frac{1}{3} < \varepsilon \][/tex]
1. Break down the inequality:
This compound inequality can be broken down into two separate inequalities:
[tex]\[ -\varepsilon < \frac{1}{x^2} - \frac{1}{3} \][/tex]
[tex]\[ \frac{1}{x^2} - \frac{1}{3} < \varepsilon \][/tex]
2. First part of the inequality:
[tex]\[ -\varepsilon < \frac{1}{x^2} - \frac{1}{3} \][/tex]
Add [tex]\(\frac{1}{3}\)[/tex] to both sides of the inequality:
[tex]\[ -\varepsilon + \frac{1}{3} < \frac{1}{x^2} \][/tex]
Inverting the inequality and simplifying:
[tex]\[ \frac{1}{x^2} > \frac{1}{3} - \varepsilon \][/tex]
Note: For [tex]\(\frac{1}{x^2} > 0\)[/tex], [tex]\( \varepsilon \)[/tex] must be such that [tex]\( \frac{1}{3} - \varepsilon > 0 \)[/tex] which means [tex]\( \varepsilon < \frac{1}{3} \)[/tex].
Therefore:
[tex]\[ x^2 < \frac{1}{ \frac{1}{3} - \varepsilon } \][/tex]
[tex]\[ x > \sqrt{ \frac{1}{ \frac{1}{3} - \varepsilon } } \][/tex]
3. Second part of the inequality:
[tex]\[ \frac{1}{x^2} - \frac{1}{3} < \varepsilon \][/tex]
Add [tex]\(\frac{1}{3}\)[/tex] to both sides of this inequality:
[tex]\[ \frac{1}{x^2} < \varepsilon + \frac{1}{3} \][/tex]
Inverting the inequality and simplifying:
[tex]\[ x^2 > \frac{1}{ \varepsilon + \frac{1}{3} } \][/tex]
[tex]\[ x < \sqrt{ \frac{1}{ \varepsilon + \frac{1}{3} } } \][/tex]
4. Combine the results:
Now, let's combine the two separate solutions:
[tex]\[ x > \sqrt{ \frac{1}{ \frac{1}{3} - \varepsilon } } \][/tex]
[tex]\[ x < \sqrt{ \frac{1}{ \varepsilon + \frac{1}{3} } } \][/tex]
Therefore, combining these two inequalities:
[tex]\[ \sqrt{ \frac{1}{ \frac{1}{3} - \varepsilon } } < x < \sqrt{ \frac{1}{ \varepsilon + \frac{1}{3} } } \][/tex]
So the solution to the inequality [tex]\(-\varepsilon<\frac{1}{x^2}-\frac{1}{3}<\varepsilon\)[/tex] is:
[tex]\[ \sqrt{ \frac{1}{ \frac{1}{3} - \varepsilon } } < x < \sqrt{ \frac{1}{ \varepsilon + \frac{1}{3} } } \][/tex]
This correct solution range for [tex]\( x \)[/tex] assumes that [tex]\( 0 < \varepsilon < \frac{1}{3} \)[/tex], so that both expressions inside the square roots remain positive.
