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
$\underset{x \rightarrow +\infty}{\lim} \ln x = +\infty$	$\underset{x \rightarrow +\infty}{\lim} e^x = +\infty$
$\underset{x \underset{>}\rightarrow 0}{\lim} \ln x = -\infty$	$\underset{x \rightarrow -\infty}{\lim} e^x = 0^+$
$\underset{x \rightarrow +\infty}{\lim} \frac {\ln x}{x} = 0^+$	$\underset{x \rightarrow +\infty}{\lim} \frac {e^x}{x} = +\infty$
$\underset{x \underset{>}\rightarrow 0}{\lim} x \ln x = 0^-$	$\underset{x \rightarrow -\infty}{\lim} xe^x = 0^-$
$\underset{x \rightarrow 0}{\lim} \frac {\ln (1+x)}{x} =1$	$\underset{x \rightarrow 0}{\lim} \frac {e^x-1}{x}=1$

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.

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

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).