Answer
(a) $x \gt 10^8$
(b) $\lim\limits_{x \to \infty}\frac{1}{\sqrt{x}} = 0$
Work Step by Step
(a) $\frac{1}{\sqrt{x}} \lt 0.0001$
$\sqrt{x} \gt \frac{1}{0.0001}$
$\sqrt{x} \gt 10^4$
$x \gt 10^8$
(b) Let $f(x) =\frac{1}{\sqrt{x}}$
This function is defined on the interval $(0, \infty)$
Let $\epsilon \gt 0$ be given.
Let $N = (\frac{1}{\epsilon})^2$
Suppose that $x \gt N$
Then:
$\vert \frac{1}{\sqrt{x}} - 0\vert \lt \vert \frac{1}{\sqrt{N}}\vert = \frac{1}{\sqrt{(\frac{1}{\epsilon})^2}} = \epsilon$
Therefore, $\lim\limits_{x \to \infty}\frac{1}{\sqrt{x}} = 0$
