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Francesco Giannino » 8.Vertical, Horizontal and Slant asymptotes


Vertical and horizontal asymptotes

Definition. The vertical line x = c is called a vertical asymptote to the graph of a function f if and only if either

\lim_{x\rightarrow c^{+}}f(x)=\pm \infty

or

\lim_{x\rightarrow c^{-}}f(x)=\pm \infty

or both.

Definition. The horizontal line y = L is a horizontal asymptote to the graph of a function f if and only if

\lim_{x\rightarrow +\infty}f(x)=L\,\,\,\text{or}\,\,\, \lim_{x\rightarrow -\infty}f(x)=L, or both.

Slant or oblique asymptotes

Definition. When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote or slant asymptote. A function f(x) is asymptotic to the straight line y = mx + q (m ≠ 0) if:

\lim_{x\rightarrow +\infty}[f(x)-(mx+q)]=0\,\,\,\text{or}\,\,\,<br />
\lim_{x\rightarrow -\infty}[f(x)-(mx+q)]=0.

In the first case the line y = mx + q is an oblique asymptote of ƒ(x) when x tends to – ∞, and in the second case the line y = mx + q is an oblique asymptote of ƒ(x) when x tends to – ∞

Slant asymptotes or oblique asymptotes

The oblique asymptote, for the function f(x), will be given by the equation y=mx+n. The value for m is computed first and is given by the following limit:

m=\lim_{x\rightarrow \pm\infty}\frac{f(x)}{x}.

It is good practice to treat the two cases (- ∞ and +∞) separately. If this limit doesn’t exist or is equal to zero, then there is no oblique asymptote in that direction.
Having m, then the value for q can be computed by

q=\lim_{x\rightarrow \pm\infty}[f(x)-mx].

If this limit fails to exist then there is no oblique asymptote in that direction, even if a limit defining m exists.

Exercises

Exercise 1. Find the vertical and horizontal asymptotes of the following function:

f(x)=\frac{4x}{x-3}.

Solution. The domain is the set of all x-values that do not give a zero in the denominator. So we set the denominator equal to zero and solve the domain:

x-3=0\Rightarrow x=3

Since we can’t have a zero in the denominator, then we can’t have x = 3. Then, the domain is: x\neq 3

Exercises

Now, we find the vertical asymptotes of the assigned function using the definition of vertical asymptote:

\lim_{x\rightarrow 3^{+}}\frac{4x}{x-3}=+\infty

\lim_{x\rightarrow 3^{-}}\frac{4x}{x-3}=-\infty

Then, the line x = 3 is the vertical asymptote.
Now, we find the horizontal asymptote of the assigned function using the definition of horizontal asymptote.

Exercises

\lim_{x\rightarrow -\infty}\frac{4x}{x-3}=

\text{Dividing both numerator and denominator by x.}=

=\lim_{x\rightarrow -\infty}\frac{4}{1-\frac{3}{x}}=4

\text{We obtain the same result when}\,\,\,x \rightarrow +\infty:

\lim_{x\rightarrow +\infty}\frac{4x}{x-3}=4

The line y = 4 is the horizontal asymptote.

Exercises

Exercise 2. Find the asymptotes of the following function:

f(x)=\frac{x^{3}-8}{x^{2}+9}.

Solution. The domain is the set of all x-values that do not give a zero in the denominator. Since in this case there are no zeroes in the denominator, then there are no forbidden x-values, and the domain is: ]-\infty,+\infty[.
Also, since there are no values forbidden to the domain, there are no vertical asymptotes.

Exercises

Now, we find the horizontal asymptote of the assigned function using the definition of horizontal asymptote:

\lim_{x\rightarrow -\infty}\frac{x^{3}-8}{x^{2}+9}=

\text{Dividing both numerator and denominator respectively by}<br />
\,\,\,x^{3}\,\,\, \text{and}\,\,\,x^{2}

$=\lim_{x\rightarrow -\infty}<br />
\frac{x^{3}(1-\frac{8}{x^{3}})}{x^{2}(1+\frac{9}{x^{2}})}=

=\lim_{x\rightarrow -\infty}<br />
\frac{x(1-\frac{8}{x^{3}})}{1+\frac{9}{x^{2}}}=-\infty.

\text{We obtain the same result when}\,\,\,x \rightarrow +\infty:

\lim_{x\rightarrow +\infty}\frac{x^{3}-8}{x^{2}+9}=+\infty

Exercises

So, there is no horizontal asymptote. This result occurs when the degree in the numerator is greater than in the denominator. Now, we can use the definition to obtain the slant asymptote:

\lim_{x\rightarrow -\infty}\frac{f(x)}{x}=<br />
\lim_{x\rightarrow -\infty}\frac{x^{3}-8}{x^{3}+9x}=1.

\text{We obtain the same result when}\,\,\,x \rightarrow +\infty:

\lim_{x\rightarrow +\infty}\frac{x^{3}-8}{x^{3}+9x}=1.

Then, the value for m is: m = 1.
Having m, then the value for q can be computed by the following limit.

Exercises

\lim_{x\rightarrow -\infty}[f(x)-mx]=<br />
\lim_{x\rightarrow -\infty}[\frac{x^{3}-8}{x^{2}+9}-x]=

=\lim_{x\rightarrow -\infty}\frac{x^{3}-8-x^{3}-9x}{x^{2}+9}=<br />
\lim_{x\rightarrow -\infty}\frac{-8-9x}{x^{2}+9}=0

\text{We obtain the same result when}\,\,\,x \rightarrow +\infty:$$<br />
$$\lim_{x\rightarrow +\infty}[\frac{x^{3}-8}{x^{2}+9}-x]=0.

Then, the value for q is: q = 0 and the line y = x the slant asymptote.

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Progetto "Campus Virtuale" dell'Università degli Studi di Napoli Federico II, realizzato con il cofinanziamento dell'Unione europea. Asse V - Società dell'informazione - Obiettivo Operativo 5.1 e-Government ed e-Inclusion