If B is an upper bound of $f(x)$ in some neighborhood of $a$, does it follow that $\lim_\limits{x\rightarrow a} f(x) \le B$?

Question

Intuitively it makes sense just from looking at a graph. If the function is bounded above by some upper bound in the neighborhood of $a$, then the limit at $a$ will naturally be bounded above by the same bound. However, I'm struggling to use the $\epsilon,\delta$ definition of a limit to actually prove this.

Answer 1

Let $b := \lim_{x\to a}f(x)$. Then for any $\epsilon>0$ there exists $\delta>0$ such that $|f(x) - b|<\epsilon$ whenever $|x-a|<\delta$, $x\neq a$. Hence, $|b| = |b-f(x)+f(x)|\le |b-f(x)|+|f(x)|\le \epsilon + B$ for $|x-a|<\delta$. Letting $\epsilon\to 0$ yields the claim.

Answer 2

Suppose the limit is $L$. Fix an arbitrary $\varepsilon>0$. The definition of $\lim_{x\to a}f(x)=L$ implies that there is a $\delta_1>0$ satisfying $$|f(x)-L|<\varepsilon\text{ for every }x\in\text{dom}[f]\text{ with }0<|x-a|<\delta_1$$ or $$L-\varepsilon<f(x)<L+\varepsilon\text{ for every }x\in\text{dom}[f]\text{ with }0<|x-a|<\delta_1$$ We're given that $B$ is an upper bound of $f$ in some neighborhood of $a$, so there's a $\delta_2>0$ satisfying $$f(x)\leq B\text{ for every }x\in(a-\delta_2,a+\delta_2)$$ It follows that for every $x\in\text{dom}[f]$, if $0<|x-a|<\min\{\delta_1,\delta_2\}$, then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$, so $$L-\varepsilon<f(x)<L+\varepsilon\text{ and }f(x)\leq B$$ and consequently $L-\varepsilon<B$.

We fixed $\varepsilon>0$ arbitrarily, so the preceding argument can be applied to any any positive real number. This implies that $L-\varepsilon<B$ is true for every $\varepsilon>0$, so $L\leq B$.