Answer
$\delta$ = $\frac{1}{8}$
Work Step by Step
The equation $\lim\limits_{x \to 2}$ $\frac{4x+1}{3x-4}$ = 4.5 is given. We are asked to find 2 values of delta. One corresponds to epsilon = 0.5 and the other to epsilon = 0.1. However if a $\delta$ works for one $\epsilon$ it will also work for all smaller values of $\epsilon$ so we must only find the corresponding $\epsilon$ to 0.5.
To do this we look back at the epsilon-delta definition of a limit.
That is, we say that where $\lim\limits_{x \to a}$ f(x) = L if, for every $\epsilon$ > 0 there exists some $\delta$ > 0 such that for every x in the domain of the function, 0 < |x - a| < $\delta$ implies that |f(x) - L| < $\epsilon$.
So we have the values of a, L, f(x), and 2 values of $\epsilon$, each of which will have its own corresponding $\delta$.
We will work backward starting with |f(x) - L| $\epsilon$ = 0.5. Substituting in known values we get | $\frac{4x+1}{3x-4}$ - 4.5| < 0.5
After simplifying we get
| $\frac{4x+1}{3x-4}$ - 4.5| < 0.5
| $\frac{4x+1}{3x-4}$ - $\frac{9}{2}$| < $\frac{1}{2}$
| $\frac{4x+1}{3x-4}$$\frac{2}{2}$ - $\frac{9}{2}$$\frac{3x-4}{3x-4}$| < $\frac{1}{2}$
| $\frac{8x+2}{6x - 8}$ - $\frac{9(3x-4)}{6x - 8}$| < $\frac{1}{2}$
| $\frac{8x+2}{6x - 8}$ - $\frac{9(3x-4)}{6x - 8}$| < $\frac{1}{2}$
| $\frac{8x+2}{6x - 8}$ - $\frac{27x-36}{6x - 8}$| < $\frac{1}{2}$
| $\frac{8x+2 - 27x + 36}{6x - 8}$| < $\frac{1}{2}$
| $\frac{38 - 19x}{6x - 8}$| < $\frac{1}{2}$
| $\frac{-19(x-2)}{2(3x - 4)}$| < $\frac{1}{2}$
$\frac{19}{2}$ |$\frac{x - 2}{3x - 4}$| < $\frac{1}{2}$
Multiplying by $\frac{2}{19}$ on both sides we get:
|$\frac{x - 2}{3x - 4}$| < $\frac{1}{2}$$\frac{2}{19}$ or just |$\frac{x - 2}{3x - 4}$| < $\frac{1}{9}$
Now we break up our fraction.
$\frac{|x - 2|}{|3x - 4|}$ < $\frac{1}{19}$
Multiply on both sides by |3x - 4|
|x - 2| < $\frac{1}{19}$ |3x - 4|
Now for a trick, we will rewrite |3x - 4| as |3x - 6 + 2|
But, |3x - 6 + 2| = |3x - 6| + |2| = |3||x - 2| + |2|
Which leaves us with:
|x - 2| < $\frac{1}{19}$ (|3||x - 2| + |2|)
Distributing and simplifying:
|x - 2| < $\frac{3}{19}$|x - 2| + $\frac{2}{19}$
|x - 2| - $\frac{3}{19}$|x - 2| < $\frac{2}{19}$
$\frac{16}{19}$|x - 2| < $\frac{2}{19}$ Now we just have to multiply through by $\frac{19}{16}$.
|x - 2| < $\frac{2}{19}$$\frac{19}{16}$
|x - 2| < $\frac{2}{16}$ Finally,
|x - 2| < $\frac{1}{8}$ This is of the form |x - a| < $\delta$ so,
So, $\delta$ = $\frac{1}{8}$
