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.
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: