Rational Functions' Asymptotes

In summary, the limit of a real one-variable function as the argument of the function approaches some real number is a specific value, which can either be an element of the range of the function or does not exist. This is determined by using rigorous definitions, such as every one-dimensional disc around L containing the image of some punctured one-dimensional disc around x0, or adjusting for limits as x increases/decreases without bound.
  • #1
syukai
5
0
First Question

If: f(x) = (x^2+1)/x

Then: f(x) = x + (1/x)

From my understanding, x would be the oblique/slant asymptote. Why is that?

Second Question

Why and how can horizontal asymptotes be crossed?
 
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  • #2
well, just glancing at it i could tell that this function has a vertical asymptote at x=0, because lim(x-->0+)f(x)=+infinity
and lim(x-->0-)f(x)=- infinity

there might be also some oblique asymptotes, let me see in a few sec.
 
  • #3
yeah, and y=x is an oblique asymptote also.
y=kx+b, is the general formula for a line. k is its slope, b- is y intercept

k=lim(x-->infinity)[f(x)/x]=1, so k=1, then

b=lim(x->infinity)[f(x)-kx]=0, so y=x is an oblique asymptote, there are no v. asymptotes.
 
  • #4
syukai said:
First Question

If: f(x) = (x^2+1)/x

Then: f(x) = x + (1/x)

From my understanding, x would be the oblique/slant asymptote. Why is that?

Second Question

Why and how can horizontal asymptotes be crossed?

Notice, according to your division result, that f(x) approaches x as x approaches negative or positive infinity, but never actually becomes x.

Another way to say this,
[tex] \[
\mathop {\lim }\limits_{x \to \pm \infty } \,f(x) = x
\]
[/tex]
 
  • #5
Actually, you may be more comfortable reading [tex] \[
\mathop {\lim }\limits_{x \to \pm \infty } \,(x + {\textstyle{1 \over x}}) = x
\]
[/tex]
 
  • #6
symbolipoint said:
Actually, you may be more comfortable reading [tex] \[
\mathop {\lim }\limits_{x \to \pm \infty } \,(x + {\textstyle{1 \over x}}) = x
\]
[/tex]
What the heck!
Are u sure?
 
  • #7
He would be more comfortable reading [tex]\lim_{x\to \infty} \frac{ x+ \frac{1}{x}}{x} = 1[/tex], or in Landau notation, [tex]x+\frac{1}{x}[/tex] ~ [tex]x[/tex], but [tex] \[
\mathop {\lim }\limits_{x \to \pm \infty } \,(x + {\textstyle{1 \over x}}) = x
\]
[/tex]

is wrong.
 
  • #8
o_o ...I'm sorry, I'm not all too familiar with limits. -_-;
 
  • #9
syukai said:
o_o ...I'm sorry, I'm not all too familiar with limits. -_-;

Well, it is kind of hard to deal with asymptotes if you are not familiar with limits. because saying in other words, asymptotes are merely some straight lines that are said to touch the graph of a function at a point at infinity, so to find them you will always have to deal with limits.
 
  • #10
Gib Z said:
He would be more comfortable reading [tex]\lim_{x\to \infty} \frac{ x+ \frac{1}{x}}{x} = 1[/tex], or in Landau notation, [tex]x+\frac{1}{x}[/tex] ~ [tex]x[/tex], but [tex] \[
\mathop {\lim }\limits_{x \to \pm \infty } \,(x + {\textstyle{1 \over x}}) = x
\]
[/tex]

is wrong.

Syukai's original function was (x^2 + 1)/x, and he correctly obtained the division result of (x + (1/x)). That makes the slant asymtote of y=x to be obvious. As x tends to positive or negative infinity, the function approaches the line y=x
 
  • #11
Yes. But to say a limit of a function of x, as x approaches infinity, is equal to another function of x, is not correct.
 
  • #12
Gib Z said:
Yes. But to say a limit of a function of x, as x approaches infinity, is equal to another function of x, is not correct.

You should have given a solid explanation, such as can be found in the middle of the website, http://en.wikipedia.org/wiki/Asymptote , which gives some understandable steps, and includes the very same example as in the start of this thread. Apparantly, we find trouble with finding a limit of a function when the function does not approach a specific value. On the other hand, when the difference between a function and another function approaches zero as the independant variable goes to plus or minus infinity, then we have a specific value of a particular difference.

Now this is clearer. The intuitive approach shown in some book such as PreCalculus of Larson, Hostetler & Edwards is good, but did not rigorously use statements of limits. A LIMIT IS A SPECIFIC VALUE. Is that correct?
 
  • #13
The limit of a real one-variable function as the argument of the function approaches some real number is either an element of the range of the function, an element of the closure of the range of the function, or does not exist (where behaviors like increasing without bound or decreasing without bound are usually denoted by the symbols [itex]\infty[/itex] or [itex]-\infty[/itex], respectively). This is only by a rigorous definition of the limit; one example of which is that L is the limit of a real one-variable function f(x) as x approaches x0 iff every one-dimensional disc around L contains the image f(D-{x0}) of some punctured one-dimensional disc D around x0 (this is slightly more geometrically appealing than the introductory definition directly using the metric). You can adjust the definition obviously for limits as x increases/decreases without bound. This doesn't really make sense if L is a variable.
Gib Z just put your statement into more rigorous terms, which does fit into limit notation.
 
Last edited:

Related to Rational Functions' Asymptotes

1. What is an asymptote?

An asymptote is a line that a curve approaches but never touches. In the context of rational functions, asymptotes are lines that the graph of the function approaches as the input (x-value) gets closer and closer to a certain value.

2. How do you find the vertical asymptote of a rational function?

To find the vertical asymptote of a rational function, set the denominator equal to 0 and solve for x. The vertical asymptote will be the value(s) of x that make the denominator equal to 0.

3. Can a rational function have more than one vertical asymptote?

Yes, a rational function can have multiple vertical asymptotes. This occurs when there are multiple values of x that make the denominator equal to 0.

4. How do you determine the horizontal asymptote of a rational function?

The degree of the numerator and denominator of a rational function determines the horizontal asymptote. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

5. Can a rational function have a slant asymptote?

Yes, a rational function can have a slant asymptote if the degree of the numerator is exactly one more than the degree of the denominator. To find the slant asymptote, use long division to divide the numerator by the denominator. The quotient will be the equation of the slant asymptote.

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