Limit at one real point.

  1. $ x_0 \in \mathbb{R}, \; l \in \mathbb{R}$ .

    $\displaystyle \underset{x \rightarrow x_0}{\lim} f(x) = l \Longleftrightarrow \...
...t \; \vert x-x_0\vert < \delta \Longrightarrow \vert f(x)-l\vert < \varepsilon.$    

    See Figure 1.
    Figure 1: Finite limit of a function at a real point.
    \begin{figure}\centerline
\mbox{
\subfigure[]{\epsfig{file=FiniteLimFinPt.eps,he...
...gure[]{\epsfig{file=FiniteLimFinPt-rectangle.eps,height=3.5cm}}
}\end{figure}
    Let us comment Figure 1(c). The (small) positive number $ \varepsilon$ determines the red band; for any value of this $ \varepsilon$ , you can find a value of $ \delta$ which determines a vertical band, in green, so that the graph of $ f$ passes through the rectangle determined by the two bands.

    For example, we prove that if $ f(x)=2x+3$ , then $ \underset{x \rightarrow 1}{\lim} f(x)=5$ . Take any $ \varepsilon >0$ . We look for a $ \delta >0$ such that

    $\displaystyle \vert x-1\vert<\delta \Longrightarrow \vert f(x)-5\vert<\varepsilon.$    

    We solve the inequality on the right:

    $\displaystyle \vert f(x)-5\vert$ $\displaystyle < \varepsilon$    
    $\displaystyle \vert 2x+3-5\vert$ $\displaystyle < \varepsilon$    
    $\displaystyle \vert 2x-2\vert$ $\displaystyle < \varepsilon$    
    $\displaystyle 2 \; \vert x-1\vert$ $\displaystyle < \varepsilon$    
    $\displaystyle \vert x-1\vert$ $\displaystyle < \frac 12 \; \varepsilon,$    

    thus, it suffices to choose any $ \delta >0$ such that $ \delta < \varepsilon /2$ , and we are done.
  2. $ x_0 \in \mathbb{R}\; , \; l = +\infty.$

    $\displaystyle \underset{x \rightarrow x_0}{\lim} f(x) = +\infty \Longleftrighta...
...exists \delta >0 \; \vert \; \vert x-x_0\vert < \delta \Longrightarrow f(x) >A.$    

    See Figure fig positive infinite limit of a function at a real point.
    Figure 2: Positive infinite limit of a function at a real point.
    \begin{figure}\centerline
\mbox{
\subfigure[]{\epsfig{file=InfLimFinPt.eps,heigh...
...subfigure[]{\epsfig{file=InfLimFinPt-rectangle.eps,height=4cm}}
}\end{figure}
  3. $ x_0 \in \mathbb{R}\; , \; l = -\infty.$

    $\displaystyle \underset{x \rightarrow x_0}{\lim} f(x) = -\infty \Longleftrighta...
...xists \delta >0 \; \vert \; \vert x-x_0\vert < \delta \Longrightarrow f(x) < A.$    

    Figure 3: Negative infinite limit of a function at $ -\infty $ .
    \begin{figure}\centering
\mbox{\epsfig{file=NegatInfLimFinPt.eps,height=4cm}}\end{figure}

Proposition 4.1.2   If $ f$ has a limit at $ x_0$ , this limit is unique.

Proof. This is true, either when the limit is finite or is infinite. We will develop here the proof for the finite case, the reader can do the work for the infinite case ($ +\infty$ or $ -\infty $ need the same tools). Without loss of generality, we suppose that $ x_0 \in \mathbb{R}$ .

Suppose that the function $ f$ tends to both real numbers $ l_1$ and $ l_2$ , when $ x$ is arbitrary close to $ x_0$ , i.e.

$\displaystyle \forall \varepsilon >0, \exists \delta >0 \; \vert \; \vert x-x_0\vert < \delta_1 \Longrightarrow \vert f(x)-l_1\vert < \frac {\varepsilon}{2}.$    

and

$\displaystyle \forall \varepsilon >0, \exists \delta >0 \; \vert \; \vert x-x_0\vert < \delta_2 \Longrightarrow \vert f(x)-l_2\vert < \frac {\varepsilon}{2}.$    

Denote $ \delta= \min ( \delta_1, \delta_2 )$ . We have:

$\displaystyle \vert x-x_0\vert < \delta \Longrightarrow \begin{cases}\vert f(x)...
...rac \vert{\varepsilon}{2}  f(x)-l_2\vert < \frac {\varepsilon}{2} \end{cases}$    

Thus:

$\displaystyle \vert x-x_0\vert < \delta \Longrightarrow \vert f(x)-l_1\vert + \vert f(x)-l_2\vert < \varepsilon$    

i.e.

$\displaystyle \vert x-x_0\vert < \delta \Longrightarrow \vert l_1-f(x)\vert + \vert f(x)-l_2\vert < \varepsilon$    

By the triangular inequality, we have:

$\displaystyle \vert l_1 - l_2 \vert = \vert l_1 - f(x)+f(x)-l_2\vert \leq l_1-f(x)\vert + \vert f(x)-l_2\vert < \varepsilon$    

Hence, the non negative number $ \vert l_1 - l_2 \vert$ is less than any positive number, so it must be equal to 0; this means that $ l_1=l_2$ .

$ \qedsymbol$

Noah Dana-Picard 2007-12-28