Answer
If $x \lt -9,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -3\vert \lt 0.1$
If $x \lt -19,~~~$ 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 $~~3 \lt f(x) \lt 3.1~~$ when $~~x \lt -9$
Therefore:
If $x \lt -9,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -3\vert \lt 0.1$
On the graph, we can see that $~~3 \lt f(x) \lt 3.05~~$ when $~~x \lt -19$
Therefore:
If $x \lt -19,~~~$ then $~~~ \vert \frac{1-3x}{\sqrt{x^2+1}} -3\vert \lt 0.05$
This illustrates Definition 8 for $\lim\limits_{x \to -\infty} \frac{1-3x}{\sqrt{x^2+1}} = 3$
