We say a line is the asymptote of a curve if the distance between the line and curve approaches zero as the curve (specifically the $$x$$ or $$y$$ coordinate of the points on the curve) goes to $$+\infty$$ or $$-\infty$$.

We study three types of asymptotes: (1) vertical, (2) horizontal, and (3) oblique (or inclined or slant).

In this section, we learn how to find them.

Vertical asymptotes

The curve has a vertical asymptote $$x=a$$ when:

as $$x$$ approaches $$a$$ (from the right or left or both directions), the distance between the points on the graph of $$y=f(x)$$ and the vertical line $$x=a$$ gets smaller and smaller but they never reach the line
$$x=a$$ (Figure 2).

More precisely:

Definition 1. We say $$x=a$$ is a vertical asymptote of (the graph of) the function $$y=f(x)$$, if one of the following conditions holds:
(a) $$f(x)\to+\infty$$ as $$x\to a$$ or $$x\to a^{+}$$ or $$x\to a^{-}$$
(b) $$f(x)\to-\infty$$ as $$x\to a$$ or $$x\to a^{+}$$ or $$x\to a^{-}$$

When do we have vertical asymptotes? There are two cases that we need to pay attention

1. There is a fraction $f(x)=\frac{g(x)}{h(x)}\quad\text{and}\quad\lim_{x\to a}h(x)=0$ If $\lim_{x\to a}g(x)\neq0$ then $$x=a$$ is a vertical asymptote, because it follows from Theorem [thm:Ch4-L/0] that $\lim_{x\to a}f(x)=+\infty\text{ or }-\infty.$ However, if $${\displaystyle \lim_{x\to a}h(x)=\lim_{x\to a}}g(x)=0$$, then $$f(x)$$ may or may not approach $$+\infty$$ or $$-\infty$$.
2. Logarithmic functions. For example, because $\lim_{x\to0^{+}}\ln x=-\infty$ the vertical line $$x=0$$ is a vertical asymptote of the graph of $$y=\ln x$$. See Figure 3.

A. The function is a fraction f(x) = g(x)/h(x)

A.1 Rational functions

If $$f(x)=\frac{P(x)}{Q(x)}$$ where $$P(x)$$ and $$Q(x)$$ are polynomials, you should follow these steps:

1. Factor $$P(x)$$ and $$Q(x)$$ if you can.
2. Cancel any common factor to simplify the fraction.
3. Take the denominator of the simplified fraction and set it equal to zero. All the zeros of the denominator are where there are vertical asymptotes.

Example 1

Find the vertical asymptotes (if any) of the following functions:
(a) $$f(x)=\dfrac{4-2x}{2x^{2}+x-1}$$

(b) $$g(x)=\dfrac{x^{2}+4x+3}{x^{2}+7x+12}$$

(c) $$h(x)=\dfrac{x+7}{x^{2}+1}$$

Solution

(a) Let’s factor out the numerator and denominator: $4-2x=2(2-x)$ To factor out the denominator, we can find the zeros: $$\underbrace{2}_{a}x^{2}+\underbrace{1}_{b}x+\underbrace{(-1)}_{c}=0$$

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-1\pm\sqrt{1^{2}-4\cdot2\cdot(-1)}}{2\cdot2}$
$x=-1\qquad x=-\frac{1}{2}$ [In fact, because $$a+c=b=1$$, we already knew that $$x=-1$$ is one of the zeros. See Section Ch0-Solutions-and-roots]. Therefore $\frac{4-2x}{2x^{2}+x-1}=\frac{-2(x-2)}{2(x+1)(x+\frac{1}{2})}$ We cannot simplify this fraction further. Thus, the zeros of the denominator $$x=-1$$) and $$x=-1/2$$ are the vertical asymptotes (see Figure 4).

(b) To factor out the numerator and denominator, we can use the quadratic formula ($$-b\pm\sqrt{b^{2}-4ac})/(2a)$$, or in this case to simply guess them. For the numerator, we need two numbers that their sum is 4 and their product is 3. It is easy to guess that they are 3 and thus $x^{2}+4x+3=(x+3)(x+1)$ For the denominator, we are looking for two numbers that their sum is 7 and their product is 12. Again, it is easy to guess that they are 4 and 3. Thus $x^{2}+7x+12=(x+3)(x+4)$ and $\frac{x^{2}+4x+3}{x^{2}+7x+12}=\frac{(x+3)(x+1)}{(x+4)(x+3)}=\frac{x+1}{x+4}\quad(\text{if }x\neq-3)$ The zero of the simplified fraction is $$x=-4$$. Therefore, $$x=-4$$ is the only vertical asymptote. Note that although $$x=-4$$ and $$x=-3$$ both make the denominator of the original zero, but $$x=-3$$ is not a vertical asymptote because $\lim_{x\to-3}g(x)=\lim_{x\to-3}\frac{x+1}{x+4}=\frac{-3+1}{-3+4}=-2,$ and $$x=-4$$ is a vertical asymptote because $\lim_{x\to-4^{-}}g(x)=\lim_{x\to-4^{-}}\frac{x+1}{x+4}\overset{\left[\frac{-3}{0^{-}}\right]}{=}+\infty$ and $\lim_{x\to-4^{+}}g(x)=\lim_{x\to-4^{+}}\frac{x+1}{x+4}\overset{\left[\frac{-3}{0^{+}}\right]}{=}-\infty$
[Whenever in doubt about the sign of the denominator as $$x$$ approaches $$a$$ from the left or right, test it for a number on the correct side of $$a$$. For example, as $$x\to-4^{-}$$, we test for a number on the left of $$-4$$, say $$x=-4.5$$. In this case $$x+4=-4.5+4=-0.5<0$$. And as $$x\to-4^{+}$$, we test for a number on the right of $$-4$$, say $$x=-3.9$$. In this case $$x+4=-3.9+4=0.1>0$$]

The graph of $$g$$ is shown in Figure 5.

(c) We cannot factor the denominator and numerator and the denominator is never zero, because $x^{2}+1\geq1>0$ Therefore, the graph of $$h(x)$$ does not have any vertical asymptotes. See Figure 6.

A.2 Fractions in General

We learned how to find the vertical asymptotes of rational functions. More generally, if $$f(x)$$ is a fraction, every zero (= every root) of the denominator can be potentially where a vertical asymptote occurs. In other words, if $$f(x)=\frac{g(x)}{h(x)}$$ and $$h(a)=0$$, then we should test $$x=a$$ and see if $$f(x)$$ goes to $$+\infty$$ or $$-\infty$$ as $$x$$ approaches $$a$$ (from either side).

If $$f(x)=\dfrac{g(x)}{h(x)}$$, to find the vertical asymptotes of the graph of $$f(x)$$:

1. Find the zeros of $$h(x)$$.
2. If $$h(a)=0$$ and $$g(a)\neq0$$ then $$x=a$$ is a vertical asymptote
3. If $$h(a)=0$$ and $$g(a)=0$$, then find $\lim_{x\to a^{+}}f(x),\quad\text{or}\quad\lim_{x\to a^{-}}f(x).$ If one of the above limits is $$+\infty$$ or $$-\infty$$, then $$x=a$$ is a vertical asymptote.

• In above, we assumed $$g(x)$$ and $$h(x)$$ are continuous functions on their domains. Almost all functions that we know are continuous except the greatest integer function $$y=\left\lfloor x\right\rfloor$$ (also denoted by $$y=[\![ x]\!]$$) and some piecewise-defined functions.
• Note that if $$f(x)=g(x)/h(x)$$ and $$h(a)=0$$ but $$g(a)\neq0$$, then $$x=a$$ is a vertical asymptote because as $$x$$ approaches $$a$$ (from right or left) by Theorem [thm:Ch4-L/0] $f(x)\overset{\left[\frac{g(a)(\neq0)}{0}\right]}{\longrightarrow}+\infty\text{ or }-\infty$ [Again, here we assumed $$h(x)$$ and $$g(x)$$ are continuous functions on their domains. Therefore, as $$x$$ approaches $$a$$ (from right or left), $$g(x)\to g(a)$$ and $$h(x)\to h(a)$$]

Example 2

Find the vertical asymptotes of the function given by $f(x)=\frac{x-2}{\sqrt{4-x^{2}}}.$

Solution

The denominator $$\sqrt{4-x^{2}}$$ is zero when $$x=\pm2$$, so we need to find the limit of $$f$$ as $$x$$ approaches $$-2$$ and 2. The domain of $$f$$ is the set of all $$x$$ such that the expression under the square root is nonnegative $$4-x^{2}\geq0$$ and the denominator is not zero $$4-x^{2}\neq0$$. That is \begin{aligned} Dom(f) & =\{x|\ 4-x^{2}>0\}\\ & =\{x|\ x^{2}<4\}\\ & =\{x|\ -2<x<2\}\\ & =(-2,2).\end{aligned}
[Recall that $$x^{2}<a^{2}$$ is equivalent to $$-|a|<x<|a|$$] Therefore $$x$$ can approach $$2$$ from the left and $$-2$$ only from the right. \begin{aligned} \lim_{x\to2^{-}}f(x) & =\lim_{x\to2^{-}}\frac{x-2}{\sqrt{4-x^{2}}}\\ & =\lim_{x\to2^{-}}\frac{x-2}{\sqrt{(2-x)(2+x)}}\\ & =\lim_{x\to2^{-}}\frac{-(2-x)}{\sqrt{2-x}\sqrt{2+x}}\\ & =\lim_{x\to2^{-}}\frac{-\cancel{\sqrt{2-x}}\sqrt{2-x}}{\cancel{\sqrt{2-x}}\sqrt{2+x}}\\ & =\lim_{x\to2^{-}}\frac{-\sqrt{2-x}}{\sqrt{2+x}}\\ & =-\frac{\sqrt{2-2}}{\sqrt{2+2}}\\ & =0.\end{aligned}
Therefore $$x=2$$ is not a vertical asymptote of the graph of $$f$$.
On the other hand, as $$x\to-2^{+}$$, the denominator goes to 0 through positive numbers [recall that the square root is always nonnegative] and the numerator goes to $$-4$$. Therefore by part (c) of Theorem [thm:Ch4-L/0] $$f(x)$$ goes to $$-\infty$$ [Symbolically $$\frac{-4}{0^{+}}=-\infty$$ or $$\frac{(-)}{(+)}=(-)$$]. Therefore, $$x=-2$$ is a vertical asymptote of the graph of $$f$$. Graph of $$f$$ is shown in Figure 7.

A.3 Trigonometric functions
• Recall that the graph of cosine is obtained by shifting the graph of sine to the left $$\pi/2$$, so if $$\cos\alpha=0$$, we know $$\sin\alpha\neq0$$ and if $$\sin\beta=0$$, we know $$\cos\beta\neq0$$.

Because $\tan x=\frac{\sin x}{\cos x},\quad\text{and}\quad\sec x=\frac{1}{\cos x}$ their graphs have vertical asymptotes wherever $$\cos x=0$$ (see Figure 8); that is, at $x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\dots$

 (a) (b) Figure 8. Wherever cosine is zero, tangent and secant have a vertical asymptote.

Similarly because $\cot x=\frac{\cos x}{\sin x},\quad\text{and}\quad\csc x=\frac{1}{\sin x}$ their graphs have vertical asymptotes wherever $$\sin x=0$$; that is, at $x=0,\pm\pi,\pm2\pi,\pm3\pi,\dots$

B. Logarithmic Functions

If $$f(x)=\ln(g(x))$$ and if $$\lim_{x\to a}g(x)=0$$, then $$x=a$$ is a vertical asymptote to the graph of $$f$$.

This is also true if $$g\to0$$ as $$x\to a^{+}$$ or $$x\to a^{-}$$.

• Because the input of the logarithm has to be positive, in above, we have to assume $$g(x)$$ approaches 0 through positive values as $$x\to a^{+}$$ or $$x\to a^{-}$$.

Example 3

Find the vertical asymptotes of the following functions:
(a) $$F(x)=\ln(x-2)$$
(b) $$g(x)=\ln\left(\sqrt{x^{2}-9}\right)$$.

Solution

(a) Method 1: Because $$x-2=0$$ for $$x=2$$ [If $$x>2$$, $$x-2>0$$. Therefore $$F(x)$$ is defined for $$x>2$$], the graph of $$F$$ has a vertical asymptote at $$x=2$$.
Method 2: Recall that if the graph of $$y=f(x)$$ is given, the graph of $$y=f(x-a)$$ ($$a>0$$) and is obtained by shifting the graph of $$f$$ to the right $$a$$ units. Because the graph of $$y=\ln(x-2)$$ is obtained by shifting the graph of $$y=\ln x$$ to the right 2 units, and the graph of $$y=\ln x$$ has a vertical asymptote at $$x=0$$, the graph of $$y=\ln(x-2)$$ has a vertical asymptote at $$x=2$$. See Figure 9.

(b) Because $$\sqrt{x^{2}-9}$$ is zero when $$x=\pm3$$, the graph of $$g$$ has vertical asymptotes at $$x=\pm3$$. Note that because the domain of $$g$$ is $$\{x|\ x<-3\text{ or }x>3\}=(-\infty,-3)\cup(3,\infty)$$, $$x$$ can approach $$-3$$ only from left and 3 only from the right. The graph of $$g$$ is shown in Figure 10.

• A function may have any number of vertical asymptotes or none at all. For example, $$f(x)=\tan x$$ has infinitely many vertical asymptotes at $$x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\cdots$$; $$g(x)=\frac{1}{x-1}$$ has only one asymptote $$x=1$$, and $$F(x)=x^{3}$$ has zero vertical asymptotes.

Horizontal asymptotes

We say a horizontal line is the asymptote of a curve if the distance between the points on the curve and this line gets smaller and smaller as $$x$$ goes to $$+\infty$$ or $$-\infty$$ (see Figure 11)

More precisely

Definition 2. We say the horizontal line $$y=b$$ is a horizontal asymptote of (the graph of) $$f(x)$$, if one of the following conditions holds:

(a) $$\lim_{x\to+\infty}f(x)=b$$

(b) $$\lim_{x\to-\infty}f(x)=b$$

• A function has, at most, two horizontal asymptotes, but a function can have infinitely many vertical asymptotes. For example, $$f(x)=\tan x$$ has infinitely many vertical asymptotes.

Example 4

Determine the end behavior of the graph of $y=\frac{2x^{2}-3}{x^{2}+1}.$

Solution

Because $\lim_{x\to\pm\infty}\frac{2x^{2}-3}{x^{2}+1}=\lim_{x\to\pm\infty}\frac{2x^{2}}{x^{2}}=2,$ its graph has only one asymptote, namely the line $$y=2$$, as shown in Figure 12.

Example 5

Determine the end behavior of the function $f(x)=\frac{1-x}{|x|+3}$

Solution

\begin{aligned} \lim_{x\to+\infty}f(x) & =\lim_{x\to+\infty}\frac{1-x}{|x|+3}\\ & =\lim_{x\to+\infty}\frac{1-x}{x+3}\\ & =\lim_{x\to+\infty}\frac{-x}{x}\\ & =-1\end{aligned}
since $$|x|=x$$ if $$x>0$$, while \begin{aligned} \lim_{x\to+\infty}f(x) & =\lim_{x\to+\infty}\frac{1-x}{|x|+3}\\ & =\lim_{x\to+\infty}\frac{1-x}{-x+3}\\ & =\lim_{x\to+\infty}\frac{-x}{-x}\\ & =1\end{aligned}
since $$|x|=-x$$ if $$x<0$$. Hence, $$f$$ has two horizontal asymptotes, namely $$y=-1$$ and $$y=1$$, as shown in Figure 13.

Example 6

Determine the end behavior of the function $g(x)=\frac{1}{e^{-x}+1}.$

Solution

Recall that $$\lim_{x\to+\infty}e^{-x}=0$$ (Theorem Ch4-Important-Limits-at-Infinity in Section ch4-Limits-at-infinity) \begin{aligned} \lim_{x\to+\infty}\frac{1}{e^{-x}+1} & =\frac{\lim_{x\to+\infty}1}{\lim_{x\to+\infty}e^{-x}+\lim_{x\to+\infty}1}\\ & =\frac{1}{0+1}\\ & =1.\end{aligned}
Since $$\lim_{x\to-\infty}e^{-x}=+\infty$$, $$\lim_{x\to-\infty}(e^{-x}+1)\overset{[\infty+1=\infty]}{=}\infty$$ \begin{aligned} \lim_{x\to+\infty}\frac{1}{e^{-x}+1} & =\frac{\lim_{x\to+\infty}1}{\lim_{x\to+\infty}(e^{-x}+1)}\\ & \overset{\left[\frac{1}{\infty}=0\right]}{=}0.\end{aligned}
Hence, the graph of $$g$$ has two asymptotes, namely $$y=1$$ and $$y=0$$, as shown in Figure 14.

Oblique (also known as inclined or slant) asymptotes

We say a line $$L$$ is the oblique asymptote of $$f(x)$$ if the graph of $$f(x)$$ gets closer and closer to this line as $$x\to+\infty$$ or $$x\to-\infty$$ (Figure 15).

More precisely

Definition 3. We say a line with equation $$y=mx+b$$ ($$m\neq0$$) is the oblique asymptote (of the graph) of $$f$$ if
(a) $${\displaystyle \lim_{x\to+\infty}\left[f(x)-(mx+b)\right]=0}$$ or
(b) $${\displaystyle \lim_{x\to-\infty}\left[f(x)-(mx+b)\right]=0}$$

• The distance between a point on the curve $$P\left(x,f(x)\right)$$ and the line $$y=mx+b$$ is $d=\frac{|f(x)-(mx+b)|}{\sqrt{m^{2}+1}}.$ But $$d\to0$$ as $$x\to+\infty$$ (or $$-\infty$$) if and only if $f(x)-(mx+b)\to0$ as $$x\to+\infty$$ (or $$-\infty$$).
• Consider a function of the form $f(x)=mx+b+\frac{P(x)}{Q(x)},$ where $$P(x)$$ and $$Q(x)$$ are two polynomials such that $\text{degree of }P(x)<\text{degree of }Q(x).$ As we explained in Section Ch4-rest-of-indeterminate-forms $\frac{P(x)}{Q(x)}\to0\quad\text{as }x\to\pm\infty.$ Therefore $f(x)-(mx+b)=\frac{P(x)}{Q(x)}\to0\quad\text{as }x\to\pm\infty,$ which shows the line $$y=mx+b$$ is the oblique asymptote of $$f(x)$$.

Example 7

Find the oblique asymptote of $f(x)=-3x+1+\frac{3x+7}{x^{4}-2x+2}$

Solution

Because $\lim_{x\to\pm\infty}\frac{3x+7}{x^{4}-2x+2}=\lim_{x\to\pm\infty}\frac{3x}{x^{4}}=0,$
$\lim_{x\to\pm\infty}\left[f(x)-(-3x+1)\right]=0$ and therefore, the line $$y=-3x+1$$ is the oblique asymptote of $$f(x)$$, as shown in Figure 16.

• If $f(x)=\frac{P_{1}(x)}{P_{2}(x)}$ where $$P_{1}(x)$$ and $$P_{2}(x)$$ are two polynomials such that $\text{degree of }P_{1}(x)=\text{degree of }P_{2}(x)+1$ then $$f(x)$$ has an oblique asymptote which is the quotient of dividing $$P_{1}(x)$$ by $$P_{2}(x)$$.

Example 8

Find the oblique asymptote of the graph of $f(x)=\frac{3x^{2}+5x-3}{x+1}.$

Solution

Let’s divide the numerator by the denominator:

Thus $\frac{3x^{2}+5x-3}{x+1}=3x+2+\frac{-5}{x+1}$ Because $\frac{-5}{x+1}\to0\quad\text{as}\quad x\to\pm\infty$ the line $$y=3x+2$$ is the oblique asymptote of the graph of $$f$$, as shown in Figure 17. Also notice that the graph of $$f$$ has a vertical asymptote at $$x=-1$$.

• More generally, when $f(x)=mx+b+g(x)$ where $$g(x)\to0$$ as $$x\to+\infty$$ (or $$-\infty)$$, the line $$y=mx+b$$ is an oblique asymptote of $$f(x)$$.
How to determine the oblique asymptotes

If the line $$y=mx+b$$ is the oblique asymptote of $$f(x)$$ as $$x\to+\infty$$ then $m=\lim_{x\to+\infty}\frac{f(x)}{x}$ and $b=\lim_{x\to+\infty}\left[f(x)-mx\right].$ If this line is the oblique asymptote of $$f(x)$$ as $$x\to-\infty$$, it is sufficient to replace $$+\infty$$ by $$-\infty$$ in the above equations.

• If $$y=mx+b$$ is the oblique asymptote of $$f(x)$$ as $$x\to+\infty$$, we have

\begin{align}\lim_{x\to+\infty}\left(f(x)-(mx+b)\right)=0\tag{i}\end{align}

or

\begin{align}\lim_{x\to+\infty}x\left[\frac{f(x)}{x}-m-\frac{b}{x}\right]=0.\tag{ii}\end{align}

•  Since $$\lim_{x\to+\infty}x=+\infty$$, we must have

\begin{align}\lim_{x\to+\infty}\left[\frac{f(x)}{x}-m-\frac{b {x}\right]=0\tag{iii}\end{align} otherwise, by Theorem the limit in (ii) would be infinite. Because $\lim_{x\to+\infty}\left[\frac{f(x)}{x}-m-\frac{b}{x}\right]=\lim_{x\to+\infty}\frac{f(x)}{x}-\lim_{x\to+\infty}m-\lim_{x\to+\infty}\frac{b}{x}$
and $$b/x\to0$$ as $$x\to+\infty$$ and $$m$$ is a constant, we have $\lim_{x\to+\infty}\frac{f(x)}{x}-m=0.$ Hence (iii) implies $\lim_{x\to+\infty}\frac{f(x)}{x}=m.$ After finding $$m$$, we may use (i) to obtain $$b$$: \begin{aligned} \lim_{x\to+\infty}[f(x)-mx-b] & =0\\ \lim_{x\to+\infty}[f(x)-mx]-\lim_{x\to+\infty}b & =0\\ \lim_{x\to+\infty}[f(x)-mx]-b & =0\end{aligned}
$\Rightarrow\lim_{x\to+\infty}[f(x)-mx]=b.$

Example 9

Find the oblique asymptote(s) of the graph of the function $f(x)=\sqrt{2x^{2}-1}-7x.$

Solution

The slope of the oblique asymptote as $$x\to+\infty$$ is \begin{aligned} m_{1} & =\lim_{x\to+\infty}\frac{f(x)}{x}\\ & =\lim_{x\to+\infty}\frac{\sqrt{2x^{2}-1}-7x}{x}\\ & =\lim_{x\to+\infty}\frac{\sqrt{x^{2}\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\\ & =\lim_{x\to+\infty}\frac{|x|\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\end{aligned}
Since $$|x|=x$$ when $$x>0$$, we have \begin{aligned} m_{1} & =\lim_{x\to+\infty}\frac{\bcancel{x}\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7\bcancel{x}}{\bcancel{x}}\\ & =\lim_{x\to+\infty}\left(\sqrt{2-\frac{1}{x^{2}}}-7\right)\\ & =\sqrt{2-\lim_{x\to+\infty}\frac{1}{x^{2}}}-7\\ & =\sqrt{2-0}-7\\ & =\sqrt{2}-7\end{aligned}
The $$y$$-intercept of the asymptote is
\begin{aligned} b_{1} & =\lim_{x\to+\infty}[f(x)-(\sqrt{2}-7)x]\\ & =\lim_{x\to+\infty}\left[\left(\sqrt{2x^{2}-1}-7x\right)-(\sqrt{2}-7)x\right]\\ & =\lim_{x\to+\infty}\left[\sqrt{2x^{2}-1}-\sqrt{2}x\right]\end{aligned}
Multiplying and dividing by
the conjugate of $$\sqrt{2x^{2}-1}-\sqrt{2}$$ \begin{aligned} b_{1} & =\lim_{x\to+\infty}\frac{\left(\sqrt{2x^{2}-1}-\sqrt{2}x\right)\left(\sqrt{2x^{2}-1}+\sqrt{2}x\right)}{\left(\sqrt{2x^{2}-1}+\sqrt{2}x\right)}\\ & =\lim_{x\to+\infty}\frac{(2x^{2}-1)-\left(\sqrt{2}x\right)^{2}}{\sqrt{2x^{2}-1}+\sqrt{2}x}\\ & =\lim_{x\to+\infty}\frac{-1}{\sqrt{2x^{2}-1}+\sqrt{2}x}\\ & \overset{\left[\frac{1}{\infty}\right]=0}{=}0.\end{aligned}
Hence, $$y=(\sqrt{2}-7)x+0$$ is the oblique asymptote of $$f(x)$$ as $$x\to+\infty$$.

The slope of the oblique asymptote as $$x\to-\infty$$ is obtained similarly: \begin{aligned} m_{2} & =\lim_{x\to-\infty}\frac{f(x)}{x}\\ & =\lim_{x\to-\infty}\frac{\sqrt{2x^{2}-1}-7x}{x}\\ & =\lim_{x\to-\infty}\frac{\sqrt{x^{2}\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\\ & =\lim_{x\to-\infty}\frac{|x|\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7x}{x}\end{aligned}
Since $$|x|=-x$$ when $$x<0$$, we have \begin{aligned} m_{2} & =\lim_{x\to-\infty}\frac{-\bcancel{x}\sqrt{\left(2-\frac{1}{x^{2}}\right)}-7\bcancel{x}}{\bcancel{x}}\\ & =\lim_{x\to-\infty}\left(-\sqrt{2-\frac{1}{x^{2}}}-7\right)\\ & =-\sqrt{2-\lim_{x\to-\infty}\frac{1}{x^{2}}}-7\\ & =\sqrt{2-0}-7\\ & =-\sqrt{2}-7\end{aligned}
In this case, we can show the $$y$$-intercept is also zero. That is, $$y=-(\sqrt{2}+7)x$$ is the oblique asymptote of $$f(x)$$ as $$x\to-\infty$$. The two oblique asymptotes of $$f(x)$$ are shown in Figure 18.