Exponential of a rational function.

$ f(x)=e^{\frac {x -1}{x-2}}$ .

The rational function $ x \mapsto (x-1)(x-2)$ is defined over $ \mathbb{R}-\{ 2 \}$ , so is $ f$ .

We compute four limits:

$\displaystyle \underset {x \rightarrow -\infty}{\lim} \frac {x -1}{x-2} = \underset {x \rightarrow -\infty}{\lim} \frac xx = 1$ $\displaystyle \Longrightarrow \underset {x \rightarrow -\infty}{\lim} f(x)=e^1=e$    
$\displaystyle \underset {x \rightarrow +\infty}{\lim} \frac {x -1}{x-2} = \underset {x \rightarrow +\infty}{\lim} \frac xx = 1$ $\displaystyle \Longrightarrow \underset {x \rightarrow +\infty}{\lim} f(x)=e^1=e$    
$\displaystyle \underset {x \underset{<}{\rightarrow} 2}{\lim} \frac {x -1}{x-2} = -\infty$ $\displaystyle \Longrightarrow \underset {x \underset{<}{\rightarrow} 2}{\lim} f(x)=0$    
$\displaystyle \underset {x \underset{>}{\rightarrow} 2}{\lim} \frac {x -1}{x-2} = +\infty$ $\displaystyle \Longrightarrow \underset {x \underset{>}{\rightarrow} 2}{\lim} f(x)=+\infty$    

It follows that the lines whose respective equations are $ y=e$ and $ x=2$ are asymptotes to the graph of $ f$ . Pay attention that the existence of the vertical asymptote is shown by the limit on the right at 2, but that on the left at 2 the function $ f$ has a finite limit.

As a composition of two differentiable functions, $ f$ is differentiable over $ \mathbb{R}-\{ 2 \}$ .

$\displaystyle \forall x \in \mathbb{R}-\{ 2 \}, \; f'(x)=\frac {-1}{(x-2)^2} e^{\frac {x -1}{x-2}}$    

thus $ f'(x)<0$ all over $ \mathbb{R}-\{ 2 \}$ . It follows that $ f$ is a decreasing function on $ (-\infty , 2 )$ and on $ (2,+\infty)$ .

The graph and its asymptotes are displayed in Figure 5.

Figure: The graph of $ f: \; x \mapsto e^{\frac {x -1}{x-2}}$ .

Noah Dana-Picard 2007-12-28