Logarithms and exponentials.

The logarithms and the exponentials can be defined in many ways. Precise definitions and proofs are given in Appendix I (i.e. chapter logs-exp).

Proposition 4.5.9       
Logarithms Exponentials
\bgroup\color{blue}$ \underset{x \rightarrow +\infty}{\lim} \ln x = +\infty$\egroup \bgroup\color{blue}$ \underset{x \rightarrow +\infty}{\lim} e^x = +\infty$\egroup
\bgroup\color{blue}$ \underset{x \underset{>}\rightarrow 0}{\lim} \ln x =
-\infty$\egroup \bgroup\color{blue}$ \underset{x \rightarrow -\infty}{\lim} e^x = 0^+$\egroup
\bgroup\color{blue}$ \underset{x \rightarrow +\infty}{\lim} \frac {\ln x}{x}
= 0^+$\egroup \bgroup\color{blue}$ \underset{x \rightarrow +\infty}{\lim} \frac {e^x}{x}
= +\infty$\egroup
\bgroup\color{blue}$ \underset{x \underset{>}\rightarrow
0}{\lim} x \ln x = 0^-$\egroup \bgroup\color{blue}$ \underset{x \rightarrow
-\infty}{\lim} xe^x = 0^-$\egroup
\bgroup\color{blue}$ \underset{x \rightarrow
0}{\lim} \frac {\ln (1+x)}{x} =1$\egroup \bgroup\color{blue}$ \underset{x \rightarrow
0}{\lim} \frac {e^x-1}{x}=1$\egroup

As an application, let us compute the limit, if it exists, of $ \frac
{(\ln x)^5}{x^8}$ for $ x$ arbitrary large. Note that we have here a situation of the type described above in subsection subsubsection Numerator and denominator have both an infinite limit.

$\displaystyle \frac {(\ln x)^8}{x^5} = \left( \frac {\ln x}{x^{5/8}} \right)^8 ...
...{5/8}} \right)^8 =\left( \frac 85 \cdot \frac {\ln x^{5/8}}{x^{5/8}} \right)^8.$    

As $ \underset{x \rightarrow +\infty}{\lim} \frac {\ln t}{t}=0$ and $ \underset{x \rightarrow +\infty}{\lim} x^{5/8} = +\infty$ , we have: $ \underset{x \rightarrow +\infty}{\lim} \frac {(\ln
x)^5}{x^8} =0$ .

Note that there are other ways to solve this question; one of them will appear with L'Hopital's rule (Theorem L'Hopital), another one with Taylor polynomials (v.i. Def. Taylor polynomials).

Noah Dana-Picard 2007-12-28