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Let $f(x)=1/x$. Show continuity at $x=1/2$

My work:

$$\left| {x - \frac{1}{2}} \right| < \delta \Rightarrow \left| {\frac{1}{x} - 2} \right| < \varepsilon $$

$$\left| {\frac{1}{x} - 2} \right| = \left| {\frac{2}{x}\left( {\frac{1}{2} - x} \right)} \right| = \left| {\frac{2}{x}} \right|\left| {\left( {\frac{1}{2} - x} \right)} \right|$$

Now, I'm not sure how to choose $\delta$ properly. How to do it?

José Carlos Santos
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SuperStamp
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3 Answers3

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You can divide your tasks in smaller tasks:

Task 1: $x \ge \frac{1}{2}$. Then $f(x) \le 2$, so you get rid of the $“|”$ and may find a $\delta_1$.

Task 2: Do the same for $x \lt \frac{1}{2}$, which gives you a $\delta_2$.

Then $\delta := \min\{\delta_1,\delta_2\}$ will do the job for your original problem.

Keba
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If you make sure that $\delta$ is not greater than $0.1$, then $|x-0.5|<\delta$ implies a lower bound of $0.4$ on $x$. Thus, $\dfrac2{x}$ cannot be greater than $\dfrac2{0.4}=5$. Can you take it from here?

Alraxite
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First of all, let's make sure that $|1/x|$ is not too big. So choose $\delta\le \frac{1}{4}$. Then $\frac{1}{4}\le x\le \frac{3}{4}$, and therefore in particular $0\lt \frac{1}{x}\le 4$.

Thus your expression has absolute value $\le 8\left|x-\frac{1}{2}\right|$.

If additionally we make $0\lt \delta \lt \frac{\epsilon}{8}$, then we will be OK.

Putting things together, we choose $\delta$ positive and equal to the smaller of $\frac{1}{4}$ and $\frac{8}{\epsilon}$.

André Nicolas
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  • Thanks, but how did you know how choose $\delta$ properly? What would be a "bad" $\delta$ in that case? I think the counter-example would make it easier for me to understand – SuperStamp Jan 22 '14 at 20:31
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    The only problem that arises is if the $\epsilon$ we are given is kind of big. For example, we must not come to the conclusion that $\delta=\frac{\epsilon}{100}$ is good enough. For if $\epsilon=100$, that lets $x$ be $0$, and then the inequality you used is not helpful, since $|1/x|$ can be very big. More geometrically, we first of all need to make sure that $\delta$ is small enough that we have some control ovewr the $\frac{2}{x}$ term of your expression. This issue comes up in many $\epsilon$-$\delta$ calculations. – André Nicolas Jan 22 '14 at 20:38