Let $f$ and $g$ be two real functions and $a \in \mathbb R \cup \{+\infty\}$. We provide here examples and counterexamples regarding the limits of $f$ and $g$.

If $f$ has a limit as $x$ tends to $a$ then $\vert f \vert$ also?

This is true. It is a consequence of the reverse triangle inequality $$\left\vert \vert f(x) \vert – \vert l \vert \right\vert \le \vert f(x) – l \vert$$ Hence if $\displaystyle \lim\limits_{x \to a} f(x) = l$, $\displaystyle \lim\limits_{x \to a} \vert f(x) \vert = \vert l \vert$

Is the converse of previous statement also true?

It is not. Consider the function defined by: $$\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & \frac{1}{n} & \longmapsto & -1 \text{ for } n \ge 1 \text{ integer} \\ & x & \longmapsto & 1 \text{ otherwise} \end{array}$$ $\vert f \vert$ is the constant function equal to $1$, hence $\vert f \vert$ has $1$ for limit as $x$ tends to zero. However $\lim\limits_{x \to 0} f(x)$ doesn’t exist.

If $f$ has a limit at $a$ and $g$ not, then $f+g$ doesn’t have a limit at $a$?

This is a tricky question! The case where $\displaystyle \lim\limits_{x \to a} f(x) = l$ is finite is covered in the next paragraph. If $l$ is infinite, $f+g$ can have a limit at $a$. Consider $$\begin{array}{l|rcl} g : & (0,1) & \longrightarrow & \mathbb R \\ & x & \longmapsto & \sin \frac{1}{x} \end{array}$$ and $$\begin{array}{l|rcl} f : & (0,1) & \longrightarrow & \mathbb R \\ & x & \longmapsto & \frac{1}{x} \end{array}$$ We have $\displaystyle \lim\limits_{x \to 0^+} f(x)=+\infty$, $g$ doesn’t have a right limit at $0$ but $\displaystyle \lim\limits_{x \to 0^+} (f+g)(x)=+\infty$

If $f$ has a finite limit at $a$ and $g$ not, then $f+g$ doesn’t have a finite limit at $a$?

This is true. If $f+g$ was having a finite limit at $a$, writing $g=(f+g)-f$ we see that $g$ would have a limit as the difference of two functions having a limit at a point has a limit at that point. In contradiction with our hypothesis.

If $f$ is defined on (u,v) and monotonic, then $f$ has left and right limits at all $a \in (u,v)$?

This is true. Suppose that $f$ is increasing (if $f$ is decreasing, just consider $-f$). We prove that $f$ has a left limit for all $a \in (u,v)$, the proof being similar for the right limit. Consider the set $S=\{f(x) | x < a\}$. $S$ is bounded by $f(a)$ and not empty as $f(\frac{u+a}{2}) \in S$. Therefore $S$ has a least upper bound $l$. For any $\epsilon > 0$, there exists $y < a$ with $l - \epsilon \le f(y) \le l$. As $f$ is increasing we have for $y < x < a$ $$f-\epsilon \le f(y) \le f(x) \le l$$ proving that $\displaystyle \lim\limits_{x \to a^-} f(x) = l$.

If $f$ has finite left and right limits for all $a \in [u,v]$ then $f$ is bounded on $[u,v]$?

This is true. Suppose that $(a_n)$ is a sequence of $[u,v]$ with $(f(a_n))$ unbounded. As $[u,v]$ is compact one can find a converging subsequence $(a_{\alpha_n})$ converging to $a \in [u,v]$ with $(f(a_{\alpha_n}))$ unbounded. Then either an infinite number of elements of $(a_{\alpha_n})$ are less than $a$ and $f$ cannot have a finite left limit at $a$ or an infinite number of elements of $(a_{\alpha_n})$ are larger than $a$ and $f$ cannot have a finite right limit at $a$. In both cases we get a contradiction, hence $f$ is bounded on $[u,v]$.

What about the reverse of previous statement?

It is false. Consider the Dirichlet function $$\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & 1 \text{ for } x \in \mathbb Q \\ & x & \longmapsto & 0 \text{ for } x \in \mathbb R \setminus \mathbb Q \end{array}$$ $f$ is bounded but doesn’t have a limit at any $a \in \mathbb R$ as $\mathbb Q$ is dense in $[0,1]$.

A real function $f$ has a limit at least one point $a \in \mathbb R$?

Is wrong. Consider again Dirichlet function of previous paragraph.

A function continuous at $a$ is continuous on an interval containing $a$?

Doesn’t hold. Consider $$\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & \vert x \vert \text{ for } x \in \mathbb Q \\ & x & \longmapsto & 0 \text{ for } x \in \mathbb R \setminus \mathbb Q \end{array}$$ $f$ is continuous at $0$ as we have $\vert f(x) \vert \le \vert x \vert$ for all $x \in \mathbb R$.

However for $a \neq 0$ $f$ is not continuous at $a$ as:

• if $a \in \mathbb Q$, we have $f(a) > 0$ and $f(x_n)=0$ for a sequence $(x_n)$ of irrational points converging to $a$.
• if $a \in \mathbb R \setminus \mathbb Q$, we have $f(a) = 0$ and $f(x_n) \to \vert a \vert\ > 0$ for a sequence $(x_n)$ of rational points converging to $a$.

There exists at least an interval $(u,v)$ on which $f$ is bounded?

Is wrong. Have a look to a nowhere locally bounded function.