Limits of functions and convergent sequences.

Theorem 4.3.1   Let $f$ be a function defined on a pointed neighborhood of $x_0$ . The function $f$ has a limit $l$ when $x$ is arbitrarily close to $x_0$ if, and only if, for any convergent sequence $(x_n)$ having $x_0$ as its limit, we have:

$$\underset{n \underset{>}{\rightarrow} + \infty}{\lim} f(x_n) = f(l).$$

Proof. $\qedsymbol$

Theorem 3.1 is mainly use to disprove the existence of a limit at some point, as shown in the next example.

Example 4.3.2   Let $f(x)=\sin \frac {1}{x}$ . This function has no limit at 0:

• Take $x_n=\frac {1}{\frac {\pi}{2} + 2n \pi}$ . Then $\forall n, f(x_n)=1$ and $\underset{n \underset{>}{\rightarrow} + \infty}{\lim} f(x_n) =1$ .
• Take $x_n=\frac {1}{\frac {1}{2n \pi}}$ . Then $\forall n, f(x_n)=0$ and $\underset{n \underset{>}{\rightarrow} + \infty}{\lim} f(x_n) =0$ .

As the limit of a function is unique ( v.s. Prop. 1.2 ), it cannot be equal both to 0 and to 1, therefore $f$ has no limit at 0 (in Figure 8 we show two drawings of the graph of $f$ , zooming to show how ``messy'' looks the graph for $x$ close to 0).

Figure 8: