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.