1. ## Limits (Spivak's problem)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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