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Show that the function $f(x)=1/x$ is continuous at $c>0$

I have seen techniques that restrict $\delta\leq 1/2$ and then go about showing that the function is continuous, but I am not sure how they come up with this restriction in the first place. In a similar spirit, $x^2$ is shown to be continuous at a point by restricting $\delta\leq 1.$ So my first question is that how does one come up with a restriction on $\delta?$

My second question is related more directly to the problem at hand. We know that the function $f(x)=1/x$ is strictly monotonic and bijective and therefore if at $x=c$ we take an interval around $f(x)=1/x$ as follows $(1/x-\epsilon,1/x+\epsilon).$ Then our goal is to find a $\delta>0$ such that whenever $|x-c|<\delta$ we have that $|1/x-1/c|<\epsilon.$ So I was thinking to map the points $1/x-\epsilon$ and $1/x+\epsilon$ back to their domain and so the corresponding values would be $$x_1=\frac{1}{1/x+\epsilon}$$ and $$x_2=\frac{1}{1/x-\epsilon}.$$ We see that since the function is montonic $x_1<x<x_2$ and so if we let $\delta\leq \min\{|x_1-x|,|x_2-x|\}$ then I feel that we are done. But I have not seen this being done in textbooks and so I am guessing that this wrong. Any inputs regarding this approach will be much appreciated.

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  • $\begingroup$ Your approach is incorrect: your $\delta$ should not depend on $x$. It is a real constant may or may not depend on $\epsilon$, not $x$. $\endgroup$ – user9464 Nov 19 '17 at 15:54
  • $\begingroup$ I understand that, x should be equal to c. Then would it make sense? $\endgroup$ – nls Nov 19 '17 at 16:29
  • $\begingroup$ You would need to make sure that both $x_1,x_2$ are positive or both negative as well. $\endgroup$ – user9464 Nov 19 '17 at 16:54
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I am not sure how they come up with this restriction in the first place.

Work backward.

Suppose you want to show that the function $f(x)=\frac1x$ is continuous at $x=3$. Looking at the definition, you would need to work with the inequality $$ \left|\frac1x-\frac13\right|<\epsilon, $$ which is equivalent to $$ \frac{|x-3|}{|3x|}<\epsilon\tag{1} $$

Now you see that if $x$ is "near" $3$, then $|x-3|$ can be small and $\frac{1}{|3x|}$ can not be too big (since $|3x|$ is away from $0$). Together you can "control" the whole thing by $\epsilon$.

To write down the argument, you can do a little bit of experiment. Suppose $0<|x-3|<\delta$ and you want to guarantee that (1) is true. The first thing you must ensure is that $|x|$ should be away from $0$. So you could choose $\delta<1$. Note that at this point, you have already had an estimate: $$ \frac{1}{|3x|}<\frac{1}{6}\quad\text{ for } 0<|x-3|<1\tag{2} $$ If you make $\delta<\epsilon$ as well, then $|x-3|<\epsilon$ and together with (2), you get $$ \frac{|x-3|}{|3x|}<\frac{\epsilon}{6}\leq\epsilon $$ whenever $0<|x-3|<\delta$. Now you can put things together and let $\delta=\min\{1,\epsilon\}$.


[Added:] For the continuity (say at $x=1$) of the function $g(x)=x^2$ in your first question, let me say something about what you should not do. By definition, you would end up with the inequality $$ |x^2-1|<\epsilon\tag{3} $$ Do not try to solve the inequality (3), i.e. you don't need at all to find all the $x$ such that (3) is true. And even better, you don't need the exact version of (3). Given $0<\epsilon<1$, if you let $\delta=\epsilon$, then $0<|x-1|<\delta$ implies that $$ |x^2-1|=|x-1|\cdot|x+1|\leq \epsilon\cdot(|x|+1)\leq \epsilon (2+1). $$ This yields the continuity of $g$ at $x=1$ by the following exercise:

Exercise. Let $A\in{\bf R}$. Show that the following statements are equivalent:

  • There exists some constant $C$ such that for every $\epsilon>0$, $|A|\leqslant C\epsilon$.

  • For every $\epsilon>0$, $|A|\leqslant \epsilon$.

  • For every $\epsilon$ with $0<\epsilon<1$, $|A|\leqslant \epsilon$.

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Show that $f(x)$ is continuous at $c >0.$

Need to show: $ \lim_{x \rightarrow c} f(x) =f(c).$

Consider $x$ such that

$(c/2) \lt x\lt (3/2)c,$

I.e $ |x-c| \lt c/2.$

Let $\epsilon >0$ be given.

Choose $\delta \lt \min(c/2, \epsilon c^2/2)$.

Then $|x-c| \lt \delta$ implies

$|1/x-1/c|= |\dfrac{c-x}{xc}| \lt$

$\dfrac{|x-c|}{|c^2/2|} \lt \dfrac{2\delta}{c^2} \lt \epsilon$.

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