# Spivak Ch. 5 Limits, problem 6

## Homework Statement

Suppose the functions f and g have the following property: for all ε > 0 and all x,

If $0 < \left| x-2 \right| < \sin^{2} \left( \frac{\varepsilon^{2}}{9} \right) + \varepsilon$, then $\left| f(x) - 2 \right| < \varepsilon$.

If $0 < \left| x - 2 \right| < \varepsilon^{2}$ then $\left| g(x) - 4 \right| < \varepsilon$.

For each $\varepsilon > 0$ find a $\delta > 0$ such that for all x:

ii) if $0 < \left| x-2 \right| < \delta$, then $\left| f(x)g(x) - 8 \right| < \varepsilon$.

## The Attempt at a Solution

The solution book says:

We need: $\left| f(x) - 2 \right| < min \left( 1, \frac{\varepsilon}{2 (\left| 4 \right| + 1)} \right)$ and $\left| g(x) - 4 \right| < \frac{\varepsilon} {2 (\left| 2 \right|) + 1}$.

My question is: How did they get these fractions in the solution? I've multiplied $\left| f(x) - 2 \right|$ and $\left| g(x) - 4 \right|$ to get $\left| f(x)g(x) - 4f(x) - 2g(x) + 8 \right|$. Am I going in the right direction?

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haruspex
Homework Helper
Gold Member
That g in the denominator in the first line is g(x)?
I would start by making some simplifying substitutions: u = x+2, F(u) = f(x)-2, G(u) = g(x)-4.

The denominator in the first line is 9, not g.

jbunniii
Homework Helper
Gold Member
My question is: How did they get these fractions in the solution? I've multiplied $\left| f(x) - 2 \right|$ and $\left| g(x) - 4 \right|$ to get $\left| f(x)g(x) - 4f(x) - 2g(x) + 8 \right|$. Am I going in the right direction?
You'll have to do some more algebra to get it into a useful form. Here's a suggestion for how to start:
$$f(x)g(x) - 8 = (f(x) - 2)(g(x) - 4) + 4f(x) + 2g(x) - 16$$
Now see if you can do some more manipulations on the right hand side. You want to get more terms containing (f(x) - 2) and (g(x) - 4).

Thank you guys SO much!

jbunniii, that's exactly what I was looking for! Thank you!

A note for future visitors: the solution follows directly from part (2) of the lemma. No need for the algebraic manipulations.