Chapter 13 - Section 13.4 - Limits at Infinity; Limits of Sequences - 13.4 Exercises - Page 930: 37

$x \in (-3.01, -2.99)$

Here, we have $\dfrac{1}{(x+3)^2} \gt 10,000$ This gives: $1 \gt 10,000(+3)^2$ $\implies |x+3|^2 \lt \dfrac{1}{10,000}$ $\implies |x+3| \lt \dfrac{1}{100}$ $\implies -\dfrac{1}{100} \lt x+3 \lt \dfrac{1}{100}$ or, $-3.01 \lt x \lt -2.99$ Thus, we have $x \in (-3.01, -2.99)$