Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 7

Find the values of $\delta$ that correspond to $\varepsilon = 1$ and $\varepsilon = 0.01$ for the $\lim \limits_{x \to 1} (4 + x - 3x^3) = 2$ to illustrate the definition of the precise limit.










From the definition,

if $\quad 0 < | x - 1 | < \delta \qquad \text{ then } \qquad |(4 + x - 3x^3) - 2| < \varepsilon$

For $\varepsilon = 1$

As shown in the graph, we must examine the region near the point $(1, 2)$. Notice that we can rewrite the inequality.

$| (4 + x - 3x^3) - 2 | < 1$

$1 < | 4 + x - 3x^3 | < 3$

So we need to determine the values of $x$ for which the curve $y = 4 + x - 3x^3 $ lies between the horizontal lines $y = 3$ and $y = 1$ as shown in the graph.
Then we estimate the $x$-coordinate by drawing a vertical line at the point of intersection of the curve and the line up to the $x$-axis to get its distance from where the limit approaches so that we form...

$0.76 < x < 1.18 \qquad$ then $\qquad 1 < 4 + x - 3x^3 < 3$

The interval of the $x$ coordinates $(0.76, 1.18)$ is not symmetric about $x = 1$, the distance of $x = 1$ to the left end point is $1 - 0.76 = 0.24$ while at the right is $1.18 - 1 = 0.18$.
Therefore, we can choose $\delta$ to be smaller to these numbers to ensure tha we're able to keep within the range of epsilon, let $\delta = 0.18$. Then we
can rewrite the inequalities as follows.

$|x - 1| < 0.18 \qquad$ then $ \qquad |( 4 + x - 3x^3) - 2 < 1$

$\fbox{Thus, if $x$ is within the distance of $0.18$ from $1$, we are able to keep $f(x)$ within a distance of $1$ from $2$.}$

For $\varepsilon = 0.01$,

If we change the value of epsilon $\varepsilon = 1$ to a smaller number $\varepsilon = 0.01$, then by using the same method above we find that

$|(4 + x - 3x^3) - 2| < 0.01$
$1.99 < | 4 + x - 3x^3 | < 2.01$

We can estimate the value of $x$ as

$0.9980 < x < 1.0020 \qquad$ then $\qquad 1.99 < 4 + x - 3x^3 < 2.01$

The value of $\delta$ from the right and left of $1$ is the same, $1-0.9980 = 0.002$ and $1.0020 - 1 = 0.002$

$\fbox{Thus, if $\delta$ is $0.002$, we are able to keep $f(x)$ within a distance of $1$ from $2$}$