Algebraic operations.

The following theorem looks like the corresponding theorem for sequences (v.s. 7.1). Recall that ``undeterminate'' means that there is no automatic conclusion in such a case, and that everything can happen, finite limit, infinite limit, no limit at all, etc.

Theorem 4.4.1       

\bgroup\color{blue}$ \underset{x \rightarrow x_0}{\lim} f(x)$\egroup \bgroup\color{blue}$ \underset{x \rightarrow x_0}{\lim}g(x)$\egroup \bgroup\color{blue}$ \underset{x \rightarrow x_0}{\lim}(f+g)(x)$\egroup \bgroup\color{blue}$ \underset{x \rightarrow x_0}{\lim}(fg)(x)$\egroup \bgroup\color{blue}$ \underset{x \rightarrow x_0}{\lim} \frac {f(x)}{g(x)}$\egroup \bgroup\color{blue}$ \underset{x \rightarrow x_0}{\lim}\sqrt{f(x)}$\egroup
\bgroup\color{blue}$ l_1$\egroup \bgroup\color{blue}$ l_2 \neq 0$\egroup \bgroup\color{blue}$ l_1+l_2$\egroup \bgroup\color{blue}$ l_1l_2$\egroup \bgroup\color{blue}$ \frac {l_1}{l_2}$\egroup \bgroup\color{blue}$ \sqrt{l_1}$\egroup if \bgroup\color{blue}$ l_1>0$\egroup
\bgroup\color{blue}$ l_1 \neq 0$\egroup 0 \bgroup\color{blue}$ l_1$\egroup 0 \bgroup\color{blue}$ \infty$\egroup or no limit \bgroup\color{blue}$ \sqrt{l_1}$\egroup if \bgroup\color{blue}$ l_1>0$\egroup
0 0 0 0 undeterminate
0 if all terms are non negative
\bgroup\color{blue}$ l\neq 0$\egroup \bgroup\color{blue}$ \infty$\egroup \bgroup\color{blue}$ \infty$\egroup \bgroup\color{blue}$ \infty$\egroup 0     
0 \bgroup\color{blue}$ \infty$\egroup \bgroup\color{blue}$ \infty$\egroup undeterminate 0     
\bgroup\color{blue}$ \infty$\egroup \bgroup\color{blue}$ l\neq 0$\egroup \bgroup\color{blue}$ \infty$\egroup \bgroup\color{blue}$ \infty$\egroup \bgroup\color{blue}$ \infty$\egroup     
\bgroup\color{blue}$ \infty$\egroup 0 \bgroup\color{blue}$ \infty$\egroup undeterminate
either $ \infty$ or no limit
\bgroup\color{blue}$ +\infty$\egroup if ...
\bgroup\color{blue}$ \pm \infty$\egroup \bgroup\color{blue}$ \pm \infty$\egroup \bgroup\color{blue}$ \pm \infty$\egroup \bgroup\color{blue}$ +\infty$\egroup undeterminate \bgroup\color{blue}$ +\infty$\egroup if ...
\bgroup\color{blue}$ \pm \infty$\egroup \bgroup\color{blue}$ \mp \infty$\egroup undeterminate \bgroup\color{blue}$ -\infty$\egroup undeterminate \bgroup\color{blue}$ +\infty$\egroup if ...

We give now examples of the ``undeterminate'' cases, first of all in order to show why they are call undeterminate, then in order to show how they can be worked out.

Noah Dana-Picard 2007-12-28