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Suppose I have a function $f(t) = 0$ for $t \leq 0$ and $f(t) = 2t\sin(1/t) - \cos(1/t)$ for $t > 0$. I want to show that $f$ is discontinuous at $0$ by showing that $\limsup_{t \to 0^+} f(t) = 1$ and $\liminf_{t \to 0^+} f(t) = -1.$ (as per the question.)

Now I have seen $\limsup_{n \to \infty}$ but not have seen $\limsup_{t \to 0^+}$. So I am not sure but below I have given an attempt at this problem.

Since $\lim_{t \to 0^+}2t\sin{1/t} = 0$, for $t < \epsilon$ (some very small number), we have $$ f(t) = 2t\sin(1/t) - \cos(1/t) = -\cos(1/t) $$ Now since we know $-1 \leq -\cos(1/t) \leq 1$, $\limsup_{t \to 0^+} f(t) = 1$ and $\liminf_{t \to 0^+} f(t) = -1.$

Is this plausible? Thanks.

Martin Sleziak
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  • You seem to have the right idea. You'd need to actually show that $cos(\frac{1}{t})$ attains $-1$ and $1$ on the interval $(0, \epsilon)$ for all $\epsilon > 0$ – Adam Jaffe Dec 16 '15 at 04:35
  • Do you have to do this problem using limit superior and limit inferior? – JDZ Dec 16 '15 at 05:05
  • @JDZ yeah, the question requires that I do it this way. –  Dec 16 '15 at 05:14
  • The def'n of $\lim \sup_{t\to )^+}f(t)$ is : For only positive$ t$ let $h(t)=\sup {f(u): u\in (0,t]}$ and then $\lim \sup_{t\to 0^+}$ is $\lim_{t\to 0; t>0} h(t)$. You have the right idea. Look at $f(x)$ when $x=1/\pi n $ for natural number n – DanielWainfleet Dec 16 '15 at 06:41

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There are several definition of $\limsup_{n\to\infty} x_n$. It should not be very difficult to take the definition you are used to and get from this the corresponding definition for $t\to0^+$.

One of the definitions of limit superior of a sequence is $$\limsup_{n\to\infty} x_n = \lim_{n\to\infty} \sup_{k\ge n} x_k.$$

In your situation you would get the corresponding definition in the form $$\limsup_{t\to0^+} f(t) = \lim_{t\to0^+} \sup_{s\in(0,t]} f(s).$$

Similarly as for sequences, you can get other equivalent definitions of limit superior for $t\to0^+$.

All these various notions of limit superior can be understood as special case of limit superior of a net. However, the notion of net is only taught later in more advanced courses. (You will typically encounter it in a course on general topology.) See, for example here, here or here.

For you special case you can notice that $\lim\limits_{t\to0^+} 2t\sin\frac1t =0$. Therefore $$\limsup\limits_{t\to0^+} \left(2t\sin\frac1t + \cos\frac1t\right) = \lim\limits_{t\to0^+} 2t\sin\frac1t + \limsup\limits_{t\to0^+} \cos\frac1t = \limsup\limits_{t\to0^+} \cos\frac1t = 1.$$ (We have used property analogous to the property shown here fore sequences: How to prove $\limsup (x_{n}+y_{n})=\lim x_{n}+\limsup y_{n}$?)

To see that $\limsup\limits_{t\to0^+} \cos\frac1t = 1$ we could simply observe that:

  • We have $\cos\frac1t\le1$, therefore also $\limsup\limits_{t\to0^+} \cos\frac1t \le 1$.
  • We can find a sequence $t_n$ such that $t_n\to0^+$ and $\cos\frac1{t_n}=1$. (Simply take $t_n$'s such that $\frac1{t_n}=2\pi n$, i.e., $t_n=\frac1{2\pi n}$.) Therefore $\limsup\limits_{t\to0^+} \cos\frac1t \ge 1$.
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Martin Sleziak
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