# Is $f(x)=1/x$ continual on $(0,\infty)$?

I've never ever in fact done a delta - epsilon evidence, so I assumed I would certainly attempt my hand at one. I determined to attempt it out for $f(x)=1/x$. If I recognize appropriately from the wikipedia write-up, I intend to show for any kind of $\varepsilon>0$, there exists a $\delta>0$ such that if $|x-c|<\delta$, after that $|f(x)-f(c)|<\varepsilon$.

Anyhow, I saw that I desire something like $$|f(x)-f(c)|= \left| \frac{1}{x} - \frac{1}{c} \right|=\frac{|x-c|}{|xc|}<\varepsilon .$$ So $|x-c|<|xc|\varepsilon$, which looks comparable to the reality that I desire $|x-c|<\delta$. Nonetheless, I've additionally listened to that is never ever intended to allow $\delta$ rely on $x$.

Is this the appropriate instructions?

Just how would certainly I utilize this details to locate an equivalent $\delta$ for each and every $\epsilon$?

Many thanks!

There have actually been some complex remarks pertaining to dependancy on $x$ or on $c$, so allow me attempt to place all of it with each other.

You are proper that $\delta$ needs to not rely on $x$. Nonetheless, when one is confirming that $f(x)$ is continual *at $c$ *, after that $\delta$ is permitted to rely on both $\epsilon$ *and also * $c$.

Bear in mind the definition : $f(x)$ is continual *at $c$ * if and also just for every single $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that for all $x$, if $|x-c|\lt \delta$, after that $|f(x)-f(c)|\lt \epsilon$.

Notification just how the presence of $\delta$ is stated *prior to * $x$ ever before enters into the image? That is a sign that $\delta$ can not rely on $x$. On the various other hand, both $f(x)$, $\epsilon$, and also $c$ take place *prior to * $\delta$, which suggests that, lacking any kind of sign on the contrary, $\delta$ is permitted to rely on $f(x)$ (clearly), on $\epsilon$, *and also * on $c$.

So below, you can not select $\delta=|xc|\epsilon$, since that would certainly make $\delta$ rely on $x$.

The means to navigate it is to get rid of the dependancy on $x$. The key below is that given that we are attempting to see to it every little thing functions if $x$ is "close enough" to $c$, after that we will certainly additionally have that $|x|$ will certainly be really near $|c|$. So we need to have the ability to to regulate that department by $x$ in the expression $\frac{|x-c|}{|xc|}$.

Just how? Well, if a certain $\delta_0$ jobs, after that any kind of smaller sized one will certainly function too. So we can constantly reduce $\delta$ a little bit extra if required. Will certainly the first point we can keep in mind is that we can constantly call for that $\delta$ be smaller sized than both $1$ and also than $\frac{c}{2}$ ; that is, we will certainly call for $\delta\lt\min\{1,\frac{c}{2}\}$. Why $1$? Due to the fact that after that I recognize that $c-1\lt x \lt c+1$ ; if $c-1\gt 0$, after that this suggests that $\frac{1}{c+1}\lt \frac{1}{x} \lt \frac{1}{c-1}$, so we can "control" the value of $\frac{1}{x}$. Why much less than $\frac{c}{2}$? Simply in instance $c-1\lt 0$. So allow $\mu=\min\{1,\frac{c}{2}\}$. After that we can end that $\frac{1}{x}\lt \frac{1}{c-\mu}$. (We can escape put simply $\delta\lt\frac{c}{2}$ ; limiting it to much less than $1$ is an usual technique, however, which is why I place it below).

So, by calling for that $\delta\lt\min\{1,\frac{c}{2}\}$, we assure that $\frac{1}{|x|}\lt \frac{1}{c-\mu}$ (bear in mind that we are working with $(0,\infty)$). What do we obtain by this? Well, look :
$$|f(x)-f(c)| = \left|\frac{1}{x}-\frac{1}{c}\right| = \left|\frac{c-x}{xc}\right| = |x-c|\frac{1}{c}\cdot\frac{1}{x} \leq |x-c|\frac{1}{c(c-\mu)}.$$ So if we *additionally * ask that $\delta\lt c(c-\mu)\epsilon$, after that we have :
$$|f(x)-f(c)|\leq |x-c|\frac{1}{c(c-\mu)} \lt \frac{\delta}{c(c-\mu)} \lt \frac{c(c-\mu)\epsilon}{c(c-\mu)} = \epsilon$$ which is what we desire!

So, in recap, what do we require? We require to see to it that $\delta$ is :

- Less than $1$ ;
- Less than $\frac{c}{2}$ ; (both of these to make certain $\frac{1}{x}\lt\frac{1}{c-\mu}$) ;
- Less than $c(c-\mu)\epsilon$, where $\mu=\min\{1,\frac{c}{2}\}$.

So, as an example, we can simply allow $\delta = \frac{1}{2}\min(1,\frac{c}{2},c(c-\mu)\epsilon)$, where $\mu=\min\{1,\frac{c}{2}\}$.

As a whole, if you can allow your $\delta$ depend just on $f(x)$ and also on $\epsilon$ yet not $c$, after that we claim $f(x)$ is *evenly continual *. This is a more powerful problem than connection, and also usually really valuable. $\frac{1}{x}$ is not evenly continual on $(0,\infty)$, nonetheless (though it gets on $[a,\infty)$ for any kind of $a\gt 0$).

The declaration in your first paragraph would certainly be the definition of consistent connection (I analyze what you created as "for all $\epsilon$, there exists a $\delta$ such that for all $x$ ..."). For a relevant inquiry, see Intuition for uniform continuity of a function on $\mathbb{R}$. Consistent connection indicates connection, yet is purely more powerful, and also actually, the function handy is continual on $(0,\infty)$, yet not evenly continual.

The declaration of connection would certainly be that for all $\epsilon$ and also all $x$ in the array, there exists a $\delta$ such that ... So $\delta$ relies on $\epsilon$ *and also * on $x$. With ease : the better the incline at $x$, the closer you will certainly need to reach $x$ for the function values to be close.

Your evidence looks ok. Delta can rely on epsilon and also on c in the definition of "continuity". A function is called evenly continual if you can confirm that offered epsilon, the called for value of delta relies on epsilon yet NOT c. 1/x is NOT evenly continual on (0,1) which is why you can not navigate the reality that delta relies on c. 1/x would certainly be evenly continual on various other periods nonetheless (periods where f' is bounded).

Related questions