2
$\begingroup$

So $f(x) = \dfrac{1}{x}$ for $x \ne 0 $ and $f(x) = 20$ if $x = 0$, and you need to prove that it is continuous or discontinuous at $x = 0$.

So we want to show that $\lvert f(x)-f(0) \rvert < ε$. I got so far as to show $\left\lvert\dfrac{1}x - 20 \right\rvert$ and that $x < \delta$. Now I know that the function is discontinuous, so is it okay to say that if $$ε = \frac{1}{\delta} - 20$$ then $$ \frac{1}x-20 < \frac{1}\delta - 20$$ $$ \frac{1}x < \frac{1}\delta$$ $$\delta < x$$

Which cannot happen so it is discontinuous?

$\endgroup$
  • 1
    $\begingroup$ the definition of not-continuos at $x=0$ is: $\exists\varepsilon>0\ \forall\delta>0$ such that $|x|<\delta$ but $|\frac{1}{x}-20|\ge\varepsilon$. $\endgroup$ – janmarqz Mar 21 '18 at 0:18
  • $\begingroup$ hmmm, so would i set a value for $\delta$? $\endgroup$ – zodross Mar 21 '18 at 0:23
  • $\begingroup$ no, you should find $\varepsilon$ to comply for any $\delta$ $\endgroup$ – janmarqz Mar 21 '18 at 0:24
  • $\begingroup$ so I can keep $\epsilon = \frac{1}\delta - 20$ and we know that $\frac{1}x - 20$ is greater because $\delta > x$ which implies $\frac{1}\delta < \frac{1}x$. $\endgroup$ – zodross Mar 21 '18 at 0:32
  • $\begingroup$ SCNR, just a quick remark: "you need to prove that it is continuous or discontinuous at x=0". Proof: Assume, the function is continuous at x=0. Then everything is fine. Assume the contrary. Then the statement is also true (the function is continuous or discontinuous). As there are no other cases to investigate, the proof is complete. -- The problem should be: Find out whether the function is continuous or discontinuous at x=20 and proof your result. :) $\endgroup$ – Philipp Imhof Mar 21 '18 at 5:40
0
$\begingroup$

As noted in the comments, you want to negate the definition of continuity.

So, we are in search of some $\epsilon>0$, let's take $\epsilon=1$, with the property that for any $\delta>0$, we have $|x|<\delta$ and $|1/x_*-20|\geq 1$ for some $|x_*|<\delta$.

I'll tackle the limit from the right, and leave the case when $x<0$ to you.

Fix an arbitrary $\delta>0$. If $0<x<\delta$ we know $\frac{1}{x}>\frac1\delta$, so as long as $$ \frac1\delta\geq 21 $$ we have $$ \frac{1}{x}\geq21\implies \frac{1}{x}-20\geq1 $$ and $$ \left|\frac1x-20 \right|>1 $$ for all of the $x$ with $x<\frac{1}{21}$ in our delta ball.

$\endgroup$
  • $\begingroup$ thank you for the edit it is a lot more clear now :) thank you for answering $\endgroup$ – zodross Mar 21 '18 at 0:44
  • $\begingroup$ certainly, glad to help! $\endgroup$ – qbert Mar 21 '18 at 0:45
  • $\begingroup$ I would just like to make sure of one thing: if $x < 0$ then do we use $\frac {-1}x < \frac{-1}{\delta}$ to solve? $\endgroup$ – zodross Mar 21 '18 at 1:48
0
$\begingroup$

This assumption $$ ε = \frac{1}{\delta } - 20 $$does not make your $\epsilon$ positive for $ \delta =1$ or many other numbers. How do you address that ?

$\endgroup$
  • $\begingroup$ could we take the absolute value of the expression? $\endgroup$ – zodross Mar 21 '18 at 1:09
  • $\begingroup$ @MehakWahla Your $ \epsilon $ should be independent of $\delta $ $\endgroup$ – Mohammad Riazi-Kermani Mar 21 '18 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.