On James Stewart's Calculus Early transcendental it says:
The definition of limit says that if any small interval $(L - \epsilon , L + \epsilon)$ is given around $L$, then we can find an interval $(a - \delta, a + \delta)$ around a such that $f$ maps all the points in $(a - \delta, a + \delta)$ (except possibly a) into the interval $(L - \epsilon , L + \epsilon)$.
However, ''small'' is not specific which contradicts the notion of a formal definition.
In the definition it says $\forall \epsilon$ but functions with a restricted range e.g. $\sin(x)$ it is impossible for $f$ to map all the points in $(a - \delta, a + \delta)$ (except possibly a) onto the interval $(L - \epsilon , L + \epsilon)$, for all $\epsilon$.
The definition requires $(a - \delta, a + \delta)$ to map into $(L - \varepsilon, L + \varepsilon)$, not onto it. There's no requirement for all of $(L - \varepsilon, L + \varepsilon)$ to be covered; we simply want the image of every point within $\delta$ of $a$, to be within $\varepsilon$ of $L$.
As a concrete example (simpler than $\sin$), consider the constant function $f(x) = 1$. Let's also take $a = 2$. We show that it approaches the limit $L = 1$ as $x \to 2$. For any given $\varepsilon > 0$, I'll choose $\delta = 10$. Then, $$x \in (a - \delta, a + \delta) = (-8, 12) \implies f(x) = 1 \in (L - \varepsilon, L + \varepsilon).$$ This proves that $\lim_{x \to 2} f(x) = 1$. Note that not every point in $(L - \varepsilon, L + \varepsilon)$ is in the range of $f$, but more importantly, the points around $a$ map within the interval. This is in the spirit of continuity: we need points nearby $a$ to map near to $L$; we don't really care if they take the full tour.