1. **State the problem:** Prove using the formal definition of limits that $$\lim_{x \to 0} -\frac{1}{x^2} = -\infty$$. 2. **Recall the formal definition for limits tending to negative infinity:** For every real number $M < 0$, there exists a $\delta > 0$ such that if $0 < |x - 0| < \delta$, then $$-\frac{1}{x^2} < M.$$ This means the function values become less than any negative number as $x$ approaches 0. 3. **Start the proof:** Let $M < 0$ be given. 4. **Manipulate the inequality:** We want to find $\delta$ such that for all $x$ with $0 < |x| < \delta$, $$-\frac{1}{x^2} < M.$$ Multiply both sides by $-1$ (which reverses the inequality): $$\frac{1}{x^2} > -M.$$ Since $M < 0$, $-M > 0$. 5. **Rewrite the inequality:** $$\frac{1}{x^2} > -M \implies x^2 < \frac{1}{-M}.$$ 6. **Choose $\delta$:** Set $$\delta = \sqrt{\frac{1}{-M}}.$$ 7. **Verify the condition:** For all $x$ with $0 < |x| < \delta$, we have $$x^2 < \delta^2 = \frac{1}{-M} \implies \frac{1}{x^2} > -M \implies -\frac{1}{x^2} < M.$$ 8. **Conclusion:** By the formal definition, since for every $M < 0$ we can find such a $\delta$, the limit is $$\lim_{x \to 0} -\frac{1}{x^2} = -\infty.$$