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Asymptotes

Oblique (or Slant) Asymptote

The line y=mx+n is an oblique (or slant) asymptote of the graph of a function f, if f(x) approaches mx+n as x increases or decreases without bound.

The following are examples of slant (oblique) asymptotes:

 Original function: $y=\;\frac{x^2+3x+2}{x-2}$ Slant asymptote: y=x+5 Original function: $y=\;\frac{-3x^2+2}{x-1}$ Slant asymptote: y=-3x-3

Definition of Slant (Oblique) Asymptote.

The line $\fs2\;y=mx+n$ is a slant asymptote of the function f if

$\lim_{\fs1x\to-\infty}$f(x)-(mx+n)$=0\;\;\;\;o\;\;\;\;\lim_{\fs1x\to+\infty}$f(x)-(mx+n)$=0\;$

Algorithm for oblique asymptotes.

 Step 1. If the function is defined on some neighborhood of infinity, find its limit at infinity. If this limit is proper or does not exist, we do not have an oblique asymptote at infinity. If it is improper, go to the next step. Step 2. Find $m=\lim_{\fs1x\to+\infty}\frac{f(x)}{x}$ If this limit diverges (or is zero), then f does not have an oblique asymptote at infinity. If it converges (and is not zero), go to the next step Step 3. Find $n=\lim_{\fs1x\to\infty}$\frac{f(x)}{x}-mx$$ If this limit diverges, then f does not have an oblique asymptote at infinity. If it converges, then the line y = mx + nis the (oblique) asymptote of f at infinity

Find the slant asymptote(s) of the following function: $\fs2f(x)=\sqrt{x^2+1}$

Firstly, we are going to study the behaviour of the function at -∞

$m=\lim_{\fs1x\to-\infty}\frac{\sqrt{x^2+1}}{x}=\lim_{\fs1x\to-\infty}\frac{-\sqrt{x^2+1}}{-x}=\lim_{\fs1x\to-\infty}-\sqrt{\frac{x^2+1}{(-x)^2}}=-1$ $n=\lim_{\fs1x\to-\infty}$\sqrt{x^2+1}-(-x)$=\lim_{\fs1x\to-\infty}$\frac{(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)}{(\sqrt{x^2+1}-x)}$$
$=\lim_{\fs1x\to-\infty}$\frac{x^2+1-x^2)}{(\sqrt{x^2+1}-x)}$=\lim_{\fs1x\to-\infty}$\frac{1}{(\sqrt{x^2+1}-x}$=0$

Therefore, the line $\fs2y=-x$ is a slant asymptote of the function

Secondly, we are going to study the behaviour of the function at +∞
$m=\lim_{\fs1x\to+\infty}\frac{\sqrt{x^2+1}}{x}=\lim_{\fs1x\to+\infty}\sqrt{\frac{x^2+1}{x^2}}=1$
$n=\lim_{\fs1x\to+\infty}$\sqrt{x^2+1}-x$=\lim_{\fs1x\to+\infty}$\frac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{(\sqrt{x^2+1}+x)}$$
$=\lim_{\fs1x\to+\infty}$\frac{x^2+1-x^2)}{(\sqrt{x^2+1}+x)}$=\lim_{\fs1x\to+\infty}$\frac{1}{(\sqrt{x^2+1}+x}$=0$

Therefore, the line $\fs2y=x$ is a slant asymptote of the function.

Graphically:

LOCATING SLANT (OBLIQUE) ASYMPTOTES OF RATIONAL FUNCTIONS

The rational function $F(x)=\;\frac{P(x)}{Q(x)}$, where P(x) and Q(x) have no common factors, has a slant asymptote if the degree of P(x) is one greater than the degree of Q(x).

The equation of the asymptote can be determined by setting y equal to the quotient of P(X) divided by Q(x).

Find the slant asymptote of the following function:

$f(x)=\;\frac{x^2+3x+2}{x-2}$

To find the slant asymptote, I'll do the long division:

The slant asymptote is the polynomial part of the answer, not the remainder.

slant asymptote:  y = x + 5

Find the oblique (slant) asymptote(s) of the function $f(x)=\frac{5x^3-8x^2-4}{-6x^2+2x+4}$
The number of oblique (slant) asymptotes is: