0
$\begingroup$

For some $a\in \Bbb{R}$ define a function $f_{a}(x) = \begin{cases} {x^{a}\cos{1\over x}}, & \text{if $x$ $\ne$ 0} \\[2ex] 0, & \text{if $x=0$} \end{cases}$. Hints firstly are preferred.

b. For what values of $a$ is $f_{a}(x)$ continuous at $\Bbb{R}$?

$Attempt$: $\lim_\limits{x\to 0}\cos{1\over x}$ does not exist. Hence, $\cos{1\over x}$ is not continuous at 0. In order for $f$ to be continuous at $\Bbb{R}$, $f$ has to be continuous at $0$. Let us look at $\lim_\limits{x\to 0}x^{a}\cos{1\over x}$. Where $a\ge0$ $\lim_\limits{x\to 0}x^{a}\cos{1\over x}=\lim_\limits{x\to 0}x^{a}\lim_\limits{x\to 0}\cos{1\over x}=0$ But when $a<0$, $\lim_\limits{x\to 0}x^{a}\cos{1\over x}=\lim_\limits{x\to 0}x^{a}\lim_\limits{x\to 0}\cos{1\over x}=\lim_\limits{x\to 0}{1\over x^{-a}}\lim_\limits{x\to 0}\cos{1\over x}$ does not exist. So $f$ is continuous everywhere when $a>0$.

c. For what values of $a$ is $f_{a}(x)$ differentiable at $\Bbb{R}$?

d. For what values of $a$ is $f'_{a}(x)$ continuous at $\Bbb{R}$?

$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

Hint

Part b): You are essentially correct in part $b$, but the limit argument you use is not good. In particular, there are problems with $\lim\limits_{x\rightarrow 0}x^a\lim\limits_{x\rightarrow 0}\cos\frac{1}{x}=0$ since the latter limit does not exist. Here's how you can do it precisely. Let $a>0$, then $x^a$, $\cos(x)$, and $\frac{1}{x}$ are all continuous functions except at $x=0$. Now since the composition and product of continuous functions is continuous, we know $x^a\cos\frac{1}{x}$ is continuous except potentially at $x=0$.

$$\lim\limits_{x\rightarrow 0}\left|x^a\cos\frac{1}{x}\right|\le\lim\limits_{x\rightarrow0}\left|x^a\right|=0$$

On the other hand, if $a<0$, then we can check the sequence of points in the form $x=\frac{1}{2\pi n}$, $n\in\mathbb{Z}$ to see that the function is not continuous at $0$. Additionally, for the $a=0$ case, we can say that the limit at $0$ doesn't exist by considering points of the form $x=\frac{1}{2\pi n}$, and $x=\frac{1}{\pi+2\pi n}$ where $n\in\mathbb{Z}$.

We can do parts c and d with similar arguments.

$\endgroup$
13
  • $\begingroup$ @MeitarAbarbanel Well, so when $a=0$, we are only considering $f(x)=\cos\frac{1}{x}$, so if $x_n=\frac{1}{2\pi n}$, then $f(x_n)=\cos(2\pi n)=1$ for every $x_n$. On the other hand, if $y_n=\frac{1}{\pi+2\pi n}$, then $f(y_n)=\cos(\pi+2\pi n)=-1$. This is a problem because $\lim\limits_{n\rightarrow\infty}x_n=\lim\limits_{n\rightarrow\infty}y_n=0$, so if $f(x)$ were continuous at $0$, then $\lim\limits_{n\rightarrow\infty}f(x_n)=\lim\limits_{n\rightarrow\infty}f(y_n)$, but this is not the case. Does this make sense? $\endgroup$
    – Peter Woolfitt
    Jan 27, 2015 at 12:32
  • $\begingroup$ Firstly, thank you very much. Should I infer that since Thank you. When you talk about the case $a=0$, how did you intend to show discontinuity? What should I do with those sequences? Should I infer that since $\lim_\limits{n\to \infty}{({1\over 2 \pi n})^0\cdot \cos2\pi n}=1$ and $\lim_\limits{n\to \infty}{({1\over 2\pi n})^0\cdot \cos2\pi n+\pi}=0$ then $f$ is not continuous? $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 12:43
  • $\begingroup$ I see... Saying it the way I did is admissible as well? $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 12:44
  • $\begingroup$ OMG what an annoying bug $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 12:45
  • $\begingroup$ And when it comes to c., should I be looking at the continuity of $f'$ or at limits of the form $\lim_\limits{x\to y}{f(x)-f(y)\over x-y}$? $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 12:50
0
$\begingroup$

Hint: The three questions correspond to levels of squeezing of the $\cos \frac{1}{x}$ wildness at $x=0$

$\endgroup$
5
  • $\begingroup$ It does get extremely wild, but I don't see how it interferes continuity. $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 10:21
  • $\begingroup$ With $a=0,$ ie. no squeezing, the function is definitely not continuous at $x=0$ $\endgroup$
    – Joffan
    Jan 27, 2015 at 10:27
  • $\begingroup$ $sin{1\over x}$ also goes wild but is continuous :( That is so confusing... $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 10:39
  • $\begingroup$ oh it isn't... At 0 $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 10:43
  • $\begingroup$ I edited it using you hint.. How is it? $\endgroup$
    – Meitar Abarbanel
    Jan 27, 2015 at 10:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .