Epsilon-delta continuity proof

[Mr Davis 97]

Homework Statement

Prove that $f(x) = \frac{1}{x}$ is continuous using the epsilon-delta definition of continuity.

Homework Equations

The Attempt at a Solution

We will assume that the domain of $f$ is $\mathbb{R} / \{ 0\}$. Let $x_0$ be in the domain. First, we look at $\displaystyle |f(x) - f(x_0)| = \frac{|x-x_0|}{|x||x_0|}$. The problem I now see is with the denominator. We need to get a bound for $\frac{1}{|x|}$ that does not depend on $x$. I am not exactly sure how I would go about doing this... Any tips?

 

[fresh_42]

What is the definition of continuity of $f(x)$ at $x_0\,?$ I seems you confused continuity and uniform continuity. In other words: What makes you think, that $\delta$ has to be independent of $x_0\,?$

 

[Mr Davis 97]

What is the definition of continuity of $f(x)$ at $x_0\,?$ I seems you confused continuity and uniform continuity. In other words: What makes you think, that $\delta$ has to be independent of $x_0\,?$

I don't think that $\delta$ is independent of $x_0$. I'm just trying to show that it is continuous. In fact, I haven't formally learned about uniform continuity yet.

 

[fresh_42]

I don't think that $\delta$ is independent of $x_0$. I'm just trying to show that it is continuous. In fact, I haven't formally learned about uniform continuity yet.

Well, that's the difference: $\delta$ independent of $x_0$ then uniform continuous, if not, as with $f(x)=\frac{1}{x}$ then ordinary continuity. Nevertheless we need a definition. And the bounds for $x_0$ is $x_0$ itself. It is a fixed value, and with it, $x$ cannot be very different from it. You don't have (and need) independence of $x_0$.

 

[Mr Davis 97]

The definition of continuity I'm using is that a function is continuous at $x_0$ if and only if $\forall \epsilon > 0 \exists \delta > 0 \forall x \in dom(f) [|x-x_0|< \delta \implies |f(x) - f(x_0) < \epsilon|]$.

Could we start with an easier one? Suppose I wanted to prove that $x^2$ is continuous at 2. So we first look at $|x^2 - 4| = |x+2||x-2|$. So we need to get a bound on $|x+2|$. How is this done?

 

[fresh_42]

It's sometimes helpful to start from behind. Say $|x^2-4|=|x+2||x-2| < |x+2|\cdot \delta < (\delta + 4)\cdot \delta < \varepsilon$ is what we want to get. Now search for a $\delta$ which does it, if $\varepsilon$ is given.

The same with $f(x)=\frac{1}{x}$. Write $x_0=c$ and treat it as a constant. You might eventually need to distinguish between points in $(-1,0) \cup (0,1)$ and the ones with $|x_0| > 1$.

 

[Mr Davis 97]

It's sometimes helpful to start from behind. Say $|x^2-4|=|x+2||x-2| < |x+2|\cdot \delta < (\delta + 4)\cdot \delta < \varepsilon$ is what we want to get. Now search for a $\delta$ which does it, if $\varepsilon$ is given.

The same with $f(x)=\frac{1}{x}$. Write $x_0=c$ and treat it as a constant. You might eventually need to distinguish between points in $(-1,0) \cup (0,1)$ and the ones with $|x_0| > 1$.

What if I supposed that $|x-2|<1$, and so $|x| < 3$, which means that $|x+2| < |x| + 2 < 5$? The thing is, my book does this to get a bound, but I'm not sure why. It goes on to claim that I should let $\delta = \min \{1, \epsilon / 5 \}$, which I also don't understand.

 

[fresh_42]

What if I supposed that $|x-2|<1$, and so $|x| < 3$, which means that $|x+2| < |x| + 2 < 5$? The thing is, my book does this to get a bound, but I'm not sure why. It goes on to claim that I should let $\delta = \min \{1, \epsilon / 5 \}$, which I also don't understand.

We have to get to $|x^2-4| \stackrel{(*)}{<} \varepsilon$. We know $|x-2| < \delta$. This means combined, we have to prove $|x+2|\cdot \delta < \varepsilon$. Now if $|x+2|<C$ for some constant, we're done, because then we have $|x^2-4|< C \cdot \delta$ and with $\delta := \varepsilon \cdot C^{-1}$ we get $(*)$.

The minimum value for $\delta$ is only to guarantee the inequalities with $\delta < 1$ and then $|x+2| < 5$ and then $|x^2-4| < C\cdot \delta = 5 \cdot \delta < \varepsilon$.

All that counts is: $\lim_{x \to x_0} f(x) = f(x_0)$. So for a given, fixed, but arbitrary small $\varepsilon$, a $\delta$ must be found, such that $|x-x_0| < \delta$ implies $|f(x)-f(x_0)| < \varepsilon$. Which delta is chosen and how doesn't matter. The implication must hold. If in doubt, make it smaller.

 

[Mark44]

The definition of continuity I'm using is that a function is continuous at $x_0$ if and only if $\forall \epsilon > 0 \exists \delta > 0 \forall x \in dom(f) [|x-x_0|< \delta \implies |f(x) - f(x_0) < \epsilon|]$.

Close, but not quite right. You shouldn't have absolute values around the whole last inequality. It should be $|f(x) - f(x_0)| < \epsilon$

Could we start with an easier one? Suppose I wanted to prove that $x^2$ is continuous at 2. So we first look at $|x^2 - 4| = |x+2||x-2|$. So we need to get a bound on $|x+2|$. How is this done?