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       

\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(x)^{g(x)}$\egroup
\bgroup\color{blue}$ a \neq 0$\egroup \bgroup\color{blue}$ b \neq 0$\egroup \bgroup\color{blue}$ a^b$\egroup
\bgroup\color{blue}$ a \neq 0$\egroup 0 \bgroup\color{blue}$ 1$\egroup
0 0 undeterminate
\bgroup\color{blue}$ 0< l < 1$\egroup \bgroup\color{blue}$ +\infty$\egroup 0
\bgroup\color{blue}$ l>1$\egroup \bgroup\color{blue}$ +\infty$\egroup \bgroup\color{blue}$ +\infty$\egroup
1 \bgroup\color{blue}$ +\infty$\egroup undeterminate

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

Theorem 4.4.5  

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

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

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

    By an easy substitution we show that

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


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

  2. By a similar way, we show that

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

Noah Dana-Picard 2007-12-28