# Epsilon delta application

1. Oct 26, 2009

### zeebo17

1. The problem statement, all variables and given/known data

Let $$f: \Re \rightarrow \Re$$ and $$g: \Re \rightarrow \Re$$ be functions such that
$$lim_{x \rightarrow 1} f(x)=\alpha$$
and
$$lim_{x \rightarrow 1} g(x)=\beta$$
for some $$\alpha, \beta \in \Re$$ with $$\alpha < \beta$$. Use the $$\epsilon-\delta$$ definition of a limit to prove there exists a number $$\delta >0$$
such that $$f(x)<g(x)$$ for all $$x$$ satisfying $$1- \delta < x< 1+ \delta$$.

2. Relevant equations

3. The attempt at a solution

By definition I know that there exists a
$$\delta_1 >0$$ s.t. $$\left|f(x)-\alpha \right|< \epsilon_1$$ $$\forall \left| x-\alpha \right| < \delta_1$$
and
$$\delta_2 >0$$ s.t. $$\left|g(x)-\beta \right|< \epsilon_2$$ $$\forall \left| x-\beta \right| < \delta_2$$.

Then I know that the $$1- \delta < x< 1+ \delta$$ can simplify to $$\left| x-1 \right| < \delta$$.

Perhaps I should set $$\delta = min( \delta_1, \delta_2)$$? It seems that I need to show that if I can get $$x$$ close enough to 1 ($$\left| x-1 \right| < \delta$$) then $$f(x)$$ can get close enough to $$\alpha$$ and $$g(x)$$ can get close enough to $$\beta$$, thus because $$\alpha < \beta$$ we have $$f(x) < g(x)$$. Am I on the right track? If so how would I start to prove this?

Thanks!

Last edited: Oct 26, 2009
2. Oct 26, 2009

### Tedjn

You will need to find $\delta$, but remember that $\delta$ depends on $\epsilon$. You're on the right track; what do you think you need to pick for epsilon so that f(x) is "close enough" to $\alpha$ and g(x) is "close enough" to $\beta$?