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.

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 $f:x \mapsto \sqrt{x^2+1}$ .

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

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

thus

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

Now,

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

whence

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

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.

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

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 $f:x \mapsto x^3-x$ .

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:

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

By the sandwich theorem (6.1), we have:

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

thus

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

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

$$f(x)-x=\sin x$$

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 $f:x \mapsto x+1+\sin x$ .