Answer
If $x \gt 12,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -(-3)\vert \lt 0.1$
If $x \gt 22,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -(-3)\vert \lt 0.05$
Work Step by Step
We can graph the function $f(x) = \frac{1-3x}{\sqrt{x^2+1}}$
On the graph, we can see that $~~-3 \lt f(x) \lt -2.9~~$ when $~~x \gt 12$
Therefore:
If $x \gt 12,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -(-3)\vert \lt 0.1$
On the graph, we can see that $~~-3 \lt f(x) \lt -2.95~~$ when $~~x \gt 22$
Therefore:
If $x \gt 22,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -(-3)\vert \lt 0.05$
This illustrates Definition 7 for $\lim\limits_{x \to \infty} \frac{1-3x}{\sqrt{x^2+1}} = -3$
