We have the following function:

$f(x) = \left\{ \begin{array}{l l} x^2 \cos\left(\dfrac{1}{x}\right) & \quad \text{if $x \neq 0$}\\ 0 & \quad \text{if $x=0$} \end{array} \right.$

We want to show it's continuous everywhere. How does one do this? For $x=0$ I know you can use the squeeze theorem, but for all other points in the domain, how can you show that the function is continuous?

  • $\begingroup$ Product and superposition of continuous functions. $\endgroup$ – Alexander Vigodner Sep 29 '14 at 15:56
  • 2
    $\begingroup$ Usually when such a problem is presented one already knows what the user above says. $\endgroup$ – Git Gud Sep 29 '14 at 15:56
  • 2
    $\begingroup$ $$\lim_{x\rightarrow 0+}|x^{2}\cos(\frac{1}{x})|\leq \lim_{x\rightarrow 0+}x^{2}=0$$ $\endgroup$ – TheOscillator Sep 29 '14 at 15:57
  • $\begingroup$ If $x\neq0$, then your $f$ is a nice smooth function near $x$ and continuous at $x$. Can you elaborate on what the problem actually is? $\endgroup$ – Joonas Ilmavirta Sep 29 '14 at 16:00

The following are theorems, which you should have seen proved, and should perhaps prove yourself:

  1. Constant functions are continuous everywhere.
  2. The identity function is continuous everywhere.
  3. The cosine function is continuous everywhere.
  4. If $f(x)$ and $g(x)$ are continuous at some point $p$, $f(g(x))$ is also continuous at that point.
  5. If $f(x)$ and $g(x)$ are continuous at some point $p$, then $f(x)g(x)$ is continuous at that point.
  6. If $f(x)$ and $g(x)$ are continuous at some point $p$, and $g(p)\ne 0$, then $\frac{f(x)}{g(x)}$ is continuous at $p$.

Then you put together the parts. For example, $\frac 1x$ is continuous everywhere except perhaps at $x=0$, by point 6, because it is a quotient of a constant function (point 1) and the identity function (point 2). Then by point 4, $\cos \frac 1x$ is continuous everywhere except perhaps at $x=0$, because it is a composition of the cosine function, which is continuous for all $x\ne 0$, and the function $\frac 1x$. You can fill in the rest.

  • $\begingroup$ Wait, is $\dfrac{1}{x}$ continuous or not? I seem to be getting contradictory answers.. $\endgroup$ – Dolma Sep 29 '14 at 18:37
  • $\begingroup$ @dolma Functions aren't “continuous or not”. A function is continuous at a point $p$ or not. The function $\frac1x$ is continuous at every point $p$ except $p=0$; at $0$ it is not continuous. But you said in your question that you were only interested in showing that $f(x)$ was continuous at points other than $0$, because you had a separate argument, based on the squeeze theorem, to show that $f(x)$ was continuous at $0$. $\endgroup$ – MJD Sep 29 '14 at 18:42

Hint: For rest of the points use the fact that $fg$ is continious if $f,g$ is continious.

  • $\begingroup$ But then the problem becomes: How can you show that $\cos(1/x)$ and $x^2$ are continuous? $\endgroup$ – Dolma Sep 29 '14 at 16:00
  • $\begingroup$ @Dolma: You can use the fact that all polynomial functions and trigonometric function are continious. $\endgroup$ – mesel Sep 29 '14 at 16:03

You see how to handle $x=0.$ For the rest:

  • Is $y=x^2$ continuous?
  • Aside from $x=0$, is $y=\cos(1/x)$ continuous?
  • Given the answers to these two questions, is $y=x^2 \cos(1/x)$ continuous?
  • $\begingroup$ Yes, I can see that $x^2$ and $\cos(1/x)$ (except for $x=0$) are continuous, but I don't know how to "prove" those are continuous either. $\endgroup$ – Dolma Sep 29 '14 at 16:01

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.