Many functions exhibit asymptotic behavior. Graphically, that is to say that their graph approaches some other geometric object (usually a line) as the graph of the function heads away from the area around the origin. In other words, asymptotic behavior involves limits, since limits are how we mathematically describe situations where a function approaches a value.

In college algebra, you may have learned how to locate several type of asymptotes. Calculus allows us to confirm these locations, by justifying their existence through limits.

A function $f(x)$ will have the horizontal asymptote $y=L$ if either $\lim\limits_{x\to\infty}f(x)=L$ or $\lim\limits_{x\to -\infty}f(x)=L$. Therefore, to find horizontal asymptotes, we simply evaluate the limit of the function as it approaches infinity, and again as it approaches negative infinity. A function can have at most two horizontal asymptotes, one in each direction.

**Example.** Find the horizontal asymptote(s) of $f(x)=\dfrac{3x+7}{2x-5}$.

$\lim\limits_{x\to \infty}\dfrac{3x+7}{2x-5}=\lim\limits_{x\to \infty}\dfrac {3+\frac{7}{x}}{2-\frac{5}{x}}=\dfrac{3+0}{2-0}=\dfrac32$ | First, we find the limit as $x$ approaches positive infinity. |

$\lim\limits_{x\to -\infty}\dfrac{3x+7}{2x-5}=\lim\limits_{x\to -\infty}\dfrac {3+\frac{7}{x}}{2-\frac{5}{x}}=\dfrac{3+0}{2-0}=\dfrac32$ | And we find the limit as $x$ approaches negative infinity. |

Therefore, the function has horizontal asymptote $y=\dfrac32$. | Since the two limits were identical, this function has a single horizontal asymptote. Since asymptotes are lines, they are described by equations, not just by numbers. |

A function $f(x)$ will have a vertical asymptote $x=c$ if one of the four one-sided infinite limits occurs there. To find possible locations for the vertical asymptotes, we check out the domain of the function. A function is not limited in the number of vertical asymptotes it may have.

**Example.** Find the vertical asymptote(s) of $f(x)=\dfrac{3x+7}{2x-5}$.

The domain of the function is $x\ne \dfrac52$. | In a rational function, the denominator cannot be zero. So we solve $2x-5\ne 0$ to find the domain. Since there is only one solution, there can be at most one vertical asymptote. |

We have $\lim\limits_{x\to \frac52 }\dfrac{3x+7}{2x-5} = \infty$. | We perform a one-sided limit as $x$ approaches the value not in the domain. In this example, substitution gives the form $\dfrac{14.5}{0}$, which is an indication of an infinite limit. Examining the denominator, we note that $2x-5>0$ when $x>\dfrac52$. Therefore, when approaching from the right, both the numerator and denominator are positive, so both the quotient and the limit will be positive. |

Therefore, the function has vertical asymptote $x=\dfrac52$. | Once you have confirmed a vertical asymptote, it is not necessary to check the other one-sided limit at that same point. Only if the first one-sided limit is not infinite does it become necessary to check the other one-sided limit. |

An oblique linear asymptote occurs when the graph of a function approaches a line that is neither horizontal nor vertical. A function $f(x)$ will have an oblique linear asymptote $L(x)=mx+b$ when either $\lim\limits_{x\to\infty}[f(x)-L(x)]=0$ or $\lim\limits_{x\to -\infty}[f(x)-L(x)]=0$. If a rational function has an oblique linear asymptote, it can be found by division. In other types of functions, it may be more difficult to locate the oblique linear asymptote. A function can have at most two oblique linear asymptotes. Furthermore, a function cannot have more than 2 asymptotes that are either horizontal or oblique linear, and then it can only have one of those on each side. This can be seen by the fact that the horizontal asymptote is equivalent to the asymptote $L(x)=b$.

**Example.** Find the oblique linear asymptote(s) of $f(x)=\dfrac{x^2+1}{x-3}$.

Dividing the numerator by the denominator, we find $f(x)=\dfrac{x^2+1}{x-3} =x+3+\dfrac{10}{x-3}$, | The function can be divided using either long division or synthetic division. |

so we suspect the asymptote is $L(x)=x+3$. | When $x$ approaches infinity, the denominator of the fraction $\dfrac{10}{x-3}$ becomes very large, so the fraction will head toward zero. Since the other terms will not head toward zero, we identify those other terms as part of the asymptote. |

We have $\lim\limits_{x\to\infty}\left(\dfrac{x^2+1}{x-3}-(x+3)\right)= \lim\limits_{x\to\infty}\left(x+3+\dfrac{10}{x-3}-x-3\right)=\lim\limits_{x\to\infty}\dfrac{10}{x-3}$ | So we evaluate the limit of the function minus its conjectured asymptote. |

and when $x$ approaches infinity, then the quantity $x-3$ also approaches infinity, therefore $\lim\limits_{x\to\infty}\dfrac{10}{x-3}=0$. | A handwaving argument of this style is typically used in this situation. It is tempting to justify it by reference to the Composition Limit Law, but that requires that the limit of the inner function exist, and here it does not. What is really required to justify this statement is either the Sandwich Theorem or a delta-epsilon proof (or actually, an M-epsilon proof). |

A similar argument would find $\lim\limits_{x\to\infty}\left(\dfrac{x^2+1}{x-3}-(x+3)\right)=0$. | Thus, the function approaches the same line in both directions, and has only one oblique linear asymptote. |

Therefore, the function has asymptote $L(x)=x+3$. | Since we proved the appropriate limit, the result follows. |

The end behavior of a function $f(x)$ need not be approximately linear. But the use of a function $P(x)$ to describe the end behavior of $f(x)$ is typically used only when $P(x)$ has a much simpler algebraic formulation than the original $f(x)$. The end-behavior function should meet the same asymptotic requirement that we have used for lines. In other words, the limit statement $\lim\limits_{x\to\infty}[f(x)-P(x)]=0$ should be true (or its equivalent statement as $x$ approaches negative infinity). Equivalently, we could verify that the limit statement $\lim\limits_{x\to\infty}\dfrac{f(x)}{P(x)}=1$ is true. For rational functions, the end-behavior candidate is typically found by dividing.

**Example.** Find the end behavior of the function $f(x)=\dfrac{3x^4+5}{x-2}$.

Dividing the numerator by the denominator, we have $f(x)=\dfrac{3x^4+5}{x-2}=3x^3+6x^2+12x+24+\dfrac{53}{x-2}$, | The function can be divided by using either long division or synthetic division. |

so we have the polynomial $P(x)=3x^3+6x^2+12x+24$ for our end-behavior function. | The end-behavior function is identified by removing the fractional term, since as $x$ approaches infinity, that term will approach zero. |

Then $\lim\limits_{x\to\infty}[f(x)-P(x)]=\lim\limits_{x\to\infty}\dfrac{53}{x-2}=0$. | Subtracting the two functions leaves the fractional term, and its limit is zero as $x$ approaches infinity. (This is another handwaving argument, as we had in the previous example, and it also really requires either the Sandwich Theorem or a delta-epsilon proof.) |

Therefore, the function's end-behavior is equivalent to $P(x)=3x^3+6x^2+12x+24$. | Since a rational function was involved, the limit as $x$ approaches negative infinity will be identical, so this function models both ends of our original function. |