Analyzing $f(x)={1 \over x}$

If you haven't already read my previous post, I suggest you do that now, as it lays the foundation for this article.

Recalling the definition of a limit

Given that $f(x) = {1 \over x}$, let's prove that $\lim_{x \to 4} f(x) = {1 \over 4}$.

Again, we try to find a good $\delta$, we can start off with $$\begin{align} \\ \left| {1 \over x} - {1 \over 4} \right| &< \varepsilon \\ \left| {4-x} \over {4x} \right| &< \varepsilon \\ {1 \over 4} \cdot {1 \over x} |4-x| & < \varepsilon \\ {1 \over 4} \cdot {1 \over x} |x-4| & < \varepsilon \\ |x-4| & < 4x \varepsilon \end{align}$$

Ok, we have our $|4-x|$, but, we can't have $\varepsilon$ dependant on $x$. Let's see what we can do about that $1/x$ on the left, since we have an opportunity to get rid of it on the right hand side. Knowing that we want to keep $x$ close to $4$, let's declare the following $$\begin{align} |4-x| < 1 \\ -1 < x < 1 \\ 3 < x < 5 \\ \end{align}$$ Ok, this is looking interesting, let's choose the small $x$, namely $3$, because, you know, mathematicians aren't afraid of thinking small... keeping that in mind let's take the reciprocal of this inequality $${1 \over 5 } < {1 \over x} < {1 \over 3}$$ And, there we have it, we can use $\delta = 12\varepsilon$ provided that $|x-4| < 1$, or more precisely, $\min(12\varepsilon, 1)$. Why can we say this?

• The fact that $1/x > 0$
• The fact that $1/4$ is in the interval $({1\over 5}, {1\over 3})$

Let $f(x) = {1 \over x }$. Then $\lim_{x \to 4} f(x) = {1 \over 4}$.

Using the precise definition; If $0 < |x-4| < \delta$, then $\left |f(x) - {1 \over 4} \right| < \varepsilon$

Let $\varepsilon > 0$, if $|x-4| < 1$ then $\delta = \min(12\varepsilon,1)$ and ${1\over x}< {1\over 3}$. Therefore we have $$\begin{align} \left |f(x) - {1 \over 4} \right| &= \left |{1 \over x} - {1 \over 4} \right| \\ &= \left| {4-x} \over {4x} \right| \\ &= {1 \over 4} \cdot {1 \over x} |4-x| \\ &= {1 \over 4} \cdot {1 \over x} |x-4| \\ &< {1 \over 4} \cdot {1 \over 3} |x-4| \\ &< {1 \over 4} \cdot {1 \over 3} \delta \\ &< {1 \over 12} \delta \\ &< {1 \over 12} 12 \varepsilon \\ \left |f(x) - {1 \over 4} \right| &< \varepsilon \\ \end{align}$$

Let's not lose track of our goal here. What we just proved, was that if we satisfy the conditions provided in the proof, the function will be within a certain $\varepsilon$ of $1/4$. Or, to put it another way, the output will be as close as we want it to $1/4$. You can test these out by plugging some values into $x$ and see what happens.

With a little bit of thought, we can see that is false to say $\lim_{x \to 0} {1 \over x} = 0$. Or put another way, it's not true that for every $\varepsilon > 0$, we can find a $|x-0|$ that makes $|f(x) - 0| < \varepsilon$. There is no $|f(x)|\ < \varepsilon$! Or, let's say we want to be within $1/100$ of $0$. So we have to satisfy $|x-0| < 1/100$ and $|f(x) - 0| < 1/100$. Observe $$\begin{align} \text{ Let } x &= 1/100\\ |x - 0| &= |1/100 -0| \\ &= |1/100| \\ \end{align}$$ And we have $$\begin{align}\\ \left|{1 \over x }-0\right| &= \left|{1 \over {1 \over 100}}\right| \\ &= 1 \cdot 100 = 100 \end{align}$$ As a matter of fact, the smaller we make $|x-0|$, the larger $|f(x)-0|$ becomes. We say that $f(x) \to \infty$, which reads "F of x, approaches infinity." And if we think about this for a moment, it makes perfect sense, if we keep in mind that $${1 \over x } = 1 \cdot {x^{-1}}$$ So how do we prove that $\lim_{x \to 0} {1\over x} = \infty$? Or put another way, how do we prove that the function grows without bounds, as the input approaches $0$? First, let's, define precisely what it means for a limit to approach infinity.

$\lim_{x \to a} f(x) = \infty$ means that for every positive number $M$ there is a positive number $\delta$ such that $$f(x) > M \text{ whenever } 0 < | x-a| < \delta$$

This is basically saying what we just showed previously, but in a more formal way. We can make $f(x)$ bigger than any number, whenever we make $x$ close to $a$, but not $a$ itself, hence the $0 <| x-a| < \delta$.

All right, let's prove it for $1/x$. First we need to find a $\delta$ that works. So, we want $\delta$, such that $f(x) > M$ $$f(x) > M = {1\over x } > M \text{ whenever } 0< |x-0| < \delta\\ \begin{align}\\ \implies 1 &> xM \\ \implies x &< {1 \over M} \text { whenever } |x| < \delta \end{align}$$ So now we have $\delta = {1 \over M}$

By definition $\lim_{x \to 0} {1\over x} = \infty$ means that for every positive number $M$ there is a positive number $\delta$ such that ${1\over x} > M \text{ whenever } 0 < | x-0| < \delta$

Let $M > 0$, choose $\delta = {1 \over M}$. If $0 < | x-0| < \delta$. Observe $$\begin{align} |x| &< \delta \\ x &< \delta\\ {1 \over x} &> {1 \over \delta}\\ {1 \over x} &> {1 \over {1\over M}}\\ {1 \over x} &> M \end{align}$$ Thus by definition, $\lim_{x \to 0} {1\over x} = \infty$

Ok, let's do more testing. Let's use $\delta = 1/10$, so that means we have to choose a $|x| < 1/10$. So, let's use $1/100$. Plugging it into $1/x$, remembering that $M=10$ we get $${1 \over {1\over 100}} = 1 \cdot {100 \over 1} = 100 > 10$$

I know, it seems like a lot of hand waving. But that's part of mathematics, I like to call it "creative trickery".