How to Use Epsilon-Delta Definition of Limits to Prove Inequality?

[zeebo17]

Homework Statement

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.

Homework Equations

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!

 

[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?