1. ## Limits (Spivak's problem)

Hi everybody, could anyone help me with this?:
Suppose that the function g has the following property:
$\displaystyle \forall{\epsilon>0}\wedge\forall{x}\quad 0<|x-2|<\epsilon^2\implies |g(x)-4|<\epsilon$

For each $\displaystyle \epsilon>0$ find a $\displaystyle \delta>0$ such that, for all x:
If $\displaystyle 0<|x-2|<\delta$, then $\displaystyle \left|\dfrac{1}{g(x)}-\dfrac{1}{4}\right|<\epsilon$

Well, I've started working here to find an expression for $\displaystyle \delta$ in terms of $\displaystyle \epsilon$:
$\displaystyle \left|\dfrac{1}{g(x)}-\dfrac{1}{4}\right|=\left|\dfrac{4-g(x)}{4g(x)}\right|=\dfrac{|4-g(x)|}{4|g(x)|}=\dfrac{|g(x)-4|}{4|g(x)|}$

Here I suppose I need to bound $\displaystyle g(x)$ but, how? If not, what's next?

2. ## Re: Limits (Spivak's problem)

You only have to show the result for $\displaystyle \varepsilon<1$. If $\displaystyle 0<|x-2|<\varepsilon^2$ then $\displaystyle 3\leq g(x)\leq 5$.

3. ## Re: Limits (Spivak's problem)

Originally Posted by girdav
You only have to show the result for $\displaystyle \varepsilon<1$. If $\displaystyle 0<|x-2|<\varepsilon^2$ then $\displaystyle 3\leq g(x)\leq 5$.
Why? Don't I have to do it for all $\displaystyle \epsilon>0$?

4. ## Re: Limits (Spivak's problem)

If you can show the result for $\displaystyle \varepsilon \leq 1$ and you take $\displaystyle \varepsilon>1$, then the $\displaystyle \delta$ which works for $\displaystyle \varepsilon=1$ will work for this $\displaystyle \varepsilon$.

5. ## Re: Limits (Spivak's problem)

Originally Posted by MATHNEM
Why? Don't I have to do it for all $\displaystyle \epsilon>0$?
Start off with any $\displaystyle \varepsilon > 0$ let $\displaystyle \varepsilon ' = \min \left\{ {1,\varepsilon } \right\}$

Now we have $\displaystyle 3<g(x)$ or $\displaystyle \frac{1}{4g(x)}< \frac{1}{12}$.

Now you can pick $\displaystyle \delta = 144(\varepsilon ')^2$

So because $\displaystyle \varepsilon' \le \varepsilon$ you have done it for any $\displaystyle \varepsilon$.