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:

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

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

$$\underset {x \underset{<}{\rightarrow} 2}{\lim} \frac {x -1}{x-2} = -\infty \Longrightarrow \underset {x \underset{<}{\rightarrow} 2}{\lim} f(x)=0$$

$$\underset {x \underset{>}{\rightarrow} 2}{\lim} \frac {x -1}{x-2} = +\infty \Longrightarrow \underset {x \underset{>}{\rightarrow} 2}{\lim} f(x)=+\infty$$

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 $\mathbb{R}-\{ 2 \}$ .

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

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