Powers.

Now take two functions $f$ and $g$ defined over the same neighborhood $\mathcal{V}$ of the real number $x_0$ (resp. of $+\infty$ , resp. of $-\infty$ ). Consider the function $h$ defined over $\mathcal{V}$ by $h(x)=f(x)^{g(x)}$ . The different cases appearing when computing limits for $h$ at $x_0$ are displayed in the following table:

Theorem 4.4.4       

$\underset{x \rightarrow x_0}{\lim} f(x)$	$\underset{x \rightarrow x_0}{\lim}g(x)$	$\underset{x \rightarrow x_0}{\lim}f(x)^{g(x)}$
$a \neq 0$	$b \neq 0$	$a^b$
$a \neq 0$	0	$1$
0	0	undeterminate
$0< l < 1$	$+\infty$	0
$l>1$	$+\infty$	$+\infty$
1	$+\infty$	undeterminate

One of the most important cases to remember is the following:

Theorem 4.4.5  

$$\underset{x \rightarrow +\infty}{\lim} \left( 1+ \frac 1x \right)^x=e.$$

Using this theorem and the algebraic properties we studied previously, many exercises are now solvable.

Example 4.4.6       

1. Compute the limit at $+\infty$ of $f(x)=(1+2/x)^x$ .

   Solution: we have

   $$f(x)=\left( 1+ \frac 2x \right) ^x =\left( 1+ \frac 2x \right)^{2... ...t]^2=. \left[ \left( 1+ \frac {1}{\frac 2x} \right)^{\; \frac x2 \; } \right]^2$$

   
   
   By an easy substitution we show that

   $$\underset{x \rightarrow +\infty}{\lim} \left( 1+ \frac {1}{\frac 2x} \right)^{\; \frac x2 \; }=e,$$

   
   
   whence

   $$\underset{x \rightarrow +\infty}{\lim} f(x)=e^2.$$

   
   

2. By a similar way, we show that

   $$\underset{x \rightarrow +\infty}{\lim} \left( 1+\frac {1}{\sqrt{x}} \right)^{\sqrt x} = e^{\sqrt{x}}.$$