Other asymptotes.

    

Definition 4.8.1   Let $ f$ be a function defined on a neighborhood of $ +\infty$ and let $ L$ be the line with equation $ y=ax+b$ . The line $ L$ is an asymptote to the graph of $ f$ if $ \underset{x \rightarrow +\infty}{\lim} [f(x)-(ax+b)]=0$ .

There is a similar definition for a neighborhood of $ -\infty $ . Please write it down.

Example 4.8.2   Let $ f(x)=2x+\frac 1x$ . As $ \underset{x \rightarrow +\infty}{\lim} [f(x)-2x]=
\underset{x \rightarrow +\infty}{\lim}(\frac 1x)=0 $ , the graph of $ f$ has an oblique asymptote whose equation is $ y=2x$ .

You can check that the $ y-$ axis is a vertical asymptote to this graph, according to def vert asympt.

Figure 13: A curve with an oblique asymptote.
\begin{figure}\mbox{\epsfig{file=hyperbola2.eps,height=5cm}}\end{figure}

We will now see an algorithm to check whether the graph of a given function has an oblique asymptote or not. WLOG, we suppose that $ \underset{x\rightarrow +\infty}{\lim} f(x)=+\infty$ .

  1. We compute $ \underset{x \rightarrow +\infty}{\lim} \frac {f(x)}{x}$ .
    1. If it is infinite, there is no oblique asymptote, and we are done.
    2. Otherwise, we denote by $ a$ this limit and proceed to the next step.
  2. We compute now $ \underset{x \rightarrow +\infty}{\lim} [f(x)-ax]$ .
    1. If it is infinite, there is no oblique asymptote, and we are done.
    2. Otherwise, we denote by $ b$ this limit and the line whose equation is $ y=ax+b$ is a an asymptote to the graph of the function.

Example 4.8.3   Take $ f(x)=\sqrt{x^2+1}$ .

On Figure 14, we display the graph of $ f$ .

Figure: The graph of $ f:x \mapsto \sqrt{x^2+1}$ .
\begin{figure}\centerline
\mbox{\epsfig{file=GraphObliqAsympt.eps,height=6cm}}\end{figure}

We work according to the algorithm above. Suppose that $ x>0$ ; then:

$\displaystyle \frac {f(x)}{x}=\frac {\sqrt{x^2+1}}{x}=\frac {x \; \sqrt{1+\frac {1}{x^2}}}{x} =\sqrt{1+\frac {1}{x^2}},$    

thus

$\displaystyle \underset{x \rightarrow +\infty}{\lim} \frac {f(x)}{x}= \underset{x \rightarrow +\infty}{\lim} \sqrt{1+\frac {1}{x^2}}=1.$    

Now,

$\displaystyle f(x)-x=\sqrt{x^2+1}-x=\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x} =\frac{1}{\sqrt{x^2+1}+x},$    

whence

$\displaystyle \underset{x \rightarrow +\infty}{\lim} (f(x)-x)=0.$    

We conclude that the line whose equation is $ y=x$ is an asymptote to the graph of $ f$ .

Working in a similar way for $ x<0$ , we can show that the line whose equation is $ y=-x$ is an asymptote to the graph of $ f$ . This result is also a consequence of the fact that $ f$ is an even function, therefore its graph is symmetric about the $ y-$ axis.

Example 4.8.4   Take $ f(x)=x^3-x.$ We compute the limit of $ f(x)/x$ when $ x$ tends to infinity.

$\displaystyle \underset{\lim}{x \rightarrow + \infty} \frac {f(x)}{x}=\underset...
... + \infty} \frac{x^3}{x}=\underset{\lim}{x \rightarrow + \infty} x^2 = +\infty.$    

Therefore the graph of the function $ f$ has no oblique asymptote for $ x$ arbitrarily close to infinity. A similar computation shows that the same situation occurs in a neighborhood of $ -\infty $ .

On Figure 15, we display the graph of $ f$ .

Figure: The graph of $ f:x \mapsto x^3-x$ .
\begin{figure}\centerline
\mbox{\epsfig{file=GraphWithoutAsymptote.eps,height=5cm}}\end{figure}

Example 4.8.5   Let $ f: \; x \mapsto x+sin x$ . As the sine function is bounded over $ \mathbb{R}$ , the limit of $ f$ at $ +\infty$ is $ +\infty$ . Let us look for an oblique asymptote:

$\displaystyle \frac {f(x)}{x}=\frac {x+ \sin x}{x} = 1 + \frac {\sin x}{x}$    

By the sandwich theorem (6.1), we have:

$\displaystyle \underset{x \rightarrow +\infty}{\lim} \frac {\sin x}{x}=0$    

thus

$\displaystyle \underset{x \rightarrow +\infty}{\lim} \frac {f(x)}{x}=1.$    

If there is an oblique asymptote, its slope must be equal to 1. Now:

$\displaystyle f(x)-x=\sin x$    

and this has no limit at infinity. Similar work has to be done at $ -\infty $ . Hence, the graph of $ f$ has no oblique asymptote.
Figure: The graph of $ f:x \mapsto x+1+\sin x$ .
\begin{figure}\centerline
\mbox{\epsfig{file=xPlus1PlusSinx.eps,height=4cm}}\end{figure}

Noah Dana-Picard 2007-12-28