91Ó°ÊÓ

We know from the exercise that we want to prove the following limit equation: $$\lim _{x \rightarrow 5} \frac{1}{x^{2}}=\frac{1}{25}$$

Recall the definition of the limit states that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 5| < δ, then |f(x) - L| < ε, with f(x) = 1/x^2 and L = 1/25.

First, we rewrite the inequality in terms of f(x) - L: $$|f(x)-\frac{1}{25}| = |\frac{1}{x^2} - \frac{1}{25}| = |\frac{25-x^2}{25x^2}|$$ Now, we want to reason about values of x so that this becomes less than ε. Notice that |x^2 - 25| = |(x - 5)(x + 5)|. Since we are only concerned with x values near 5, we know that |x + 5| is going to be a positive number close to 10. Therefore, we can rewrite the inequality as: $$|\frac{25-x^2}{25x^2}| = \frac{|x-5||x+5|}{25x^2} < \frac{10|x-5|}{25x^2}$$ We want to find δ > 0 such that 0 < |x - 5| < δ implies |f(x) - L| < ε. To achieve this, we want the fraction on the right to be less than ε: $$\frac{10|x-5|}{25x^2} < ε$$ We can rewrite the inequality in terms of δ as follows: $$\frac{10\delta}{25x^2} < ε$$ $$\delta < \frac{25x^2ε}{10}$$

From our previous inequality, we have: $$\delta < \frac{25x^2ε}{10}$$ To ensure our δ > 0, we can choose it to be the minimum value that satisfies this inequality. As a result, we can set δ in terms of ε as follows: $$\delta = \frac{25x^2ε}{10}$$

We have shown that for every ε > 0, we can find a δ > 0 (in this case, δ = 25x^2ε/10) such that if 0 < |x - 5| < δ, then |f(x) - L| < ε. Thus, by the definition of a limit, we have proven that: $$\lim _{x \rightarrow 5} \frac{1}{x^{2}}=\frac{1}{25}$$