Answer
(a) $\delta = 0.01$
(b) $\lim\limits_{x \to 2^+}\frac{1}{ln~(x-1)} = \infty$
Work Step by Step
(a) $f(x) = \frac{1}{ln~(x-1)}$
On the graph, we can see that when $2 \lt x \lt 2.01$, then $\frac{1}{ln~(x-1)} \gt 100$
We can also show this algebraically:
$\frac{1}{ln~(x-1)} = 100$
$ln(x-1) = 0.01$
$x-1 = e^{0.01}$
$x = 1+e^{0.01}$
$x = 2.01005$
Thus, if $2 \lt x \lt 2.01$, then $\frac{1}{ln~(x-1)} \gt 100$
Since $2.01 = 2+\delta,$ then $\delta = 0.01$
(b) Part (a) suggests that the following limit is true:
$\lim\limits_{x \to 2^+}\frac{1}{ln~(x-1)} = \infty$
Part (a) suggests that as $x$ approaches the value of $2$ from the right side, the value of $\frac{1}{ln~(x-1)}$ grows very large.
