4.6 Limits at infinity and asymptotes  (Page 5/14)

Before proceeding, consider the graph of $f\left(x\right)=\frac{\left(3x^2+4x\right)}{\left(x+2\right)}$ shown in [link] . As $x\to \infty$ and $x\to \text{−}\infty ,$ the graph of $f$ appears almost linear. Although $f$ is certainly not a linear function, we now investigate why the graph of $f$ seems to be approaching a linear function. First, using long division of polynomials, we can write

Since $\frac{4}{\left(x+2\right)}\to 0$ as $x\to \text{±}\infty ,$ we conclude that

Therefore, the graph of $f$ approaches the line $y=3x-2$ as $x\to \text{±}\infty .$ This line is known as an oblique asymptote    for $f$ ( [link] ).

We can summarize the results of [link] to make the following conclusion regarding end behavior for rational functions. Consider a rational function

where $a_n\ne 0\text{and}{b}_m\ne 0.$

1. If the degree of the numerator is the same as the degree of the denominator $\left(n=m\right),$ then $f$ has a horizontal asymptote of $y=a_n\text{/}{b}_m$ as $x\to \text{±}\infty .$
2. If the degree of the numerator is less than the degree of the denominator $\left(n<m\right),$ then $f$ has a horizontal asymptote of $y=0$ as $x\to \text{±}\infty .$
3. If the degree of the numerator is greater than the degree of the denominator $\left(n>m\right),$ then $f$ does not have a horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the leading terms. In addition, using long division, the function can be rewritten as
   
   $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=g\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)},$
   
   where the degree of $r\left(x\right)$ is less than the degree of $q\left(x\right).$ As a result, $\underset{x\to \text{±}\infty }{\text{lim}}r\left(x\right)\text{/}q\left(x\right)=0.$ Therefore, the values of $\left[f\left(x\right)-g\left(x\right)\right]$ approach zero as $x\to \text{±}\infty .$ If the degree of $p\left(x\right)$ is exactly one more than the degree of $q\left(x\right)$ $\left(n=m+1\right),$ the function $g\left(x\right)$ is a linear function. In this case, we call $g\left(x\right)$ an oblique asymptote.
   Now let’s consider the end behavior for functions involving a radical.

Determining end behavior for a function involving a radical

Find the limits as $x\to \infty$ and $x\to \text{−}\infty$ for $f\left(x\right)=\frac{3x-2}{\sqrt{4x^2+5}}$ and describe the end behavior of $f.$

Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of $x.$ To determine the appropriate power of $x,$ consider the expression $\sqrt{4x^2+5}$ in the denominator. Since

$\sqrt{4x^2+5}\approx \sqrt{4x^2}=2\left|x\right|$

for large values of $x$ in effect $x$ appears just to the first power in the denominator. Therefore, we divide the numerator and denominator by $\left|x\right|.$ Then, using the fact that $\left|x\right|=x$ for $x>0,$ $\left|x\right|=\text{−}x$ for $x<0,$ and $\left|x\right|=\sqrt{x^2}$ for all $x,$ we calculate the limits as follows:

$\begin{array}{ccc}\hfill \underset{x\to \infty }{\text{lim}}\frac{3x-2}{\sqrt{4x^2+5}}& =\hfill & \underset{x\to \infty }{\text{lim}}\frac{\left(1\text{/}\left|x\right|\right)\left(3x-2\right)}{\left(1\text{/}\left|x\right|\right)\sqrt{4x^2+5}}\hfill \\ & =\hfill & \underset{x\to \infty }{\text{lim}}\frac{\left(1\text{/}x\right)\left(3x-2\right)}{\sqrt{\left(1\text{/}{x}^2\right)\left(4x^2+5\right)}}\hfill \\ & =\hfill & \underset{x\to \infty }{\text{lim}}\frac{3-2\text{/}x}{\sqrt{4+5\text{/}{x}^2}}=\frac{3}{\sqrt{4}}=\frac{3}{2}\hfill \\ \hfill \underset{x\to \text{−}\infty }{\text{lim}}\frac{3x-2}{\sqrt{4x^2+5}}& =\hfill & \underset{x\to \text{−}\infty }{\text{lim}}\frac{\left(1\text{/}\left|x\right|\right)\left(3x-2\right)}{\left(1\text{/}\left|x\right|\right)\sqrt{4x^2+5}}\hfill \\ & =\hfill & \underset{x\to \text{−}\infty }{\text{lim}}\frac{\left(\mathrm{-1}\text{/}x\right)\left(3x-2\right)}{\sqrt{\left(1\text{/}{x}^2\right)\left(4x^2+5\right)}}\hfill \\ & =\hfill & \underset{x\to \text{−}\infty }{\text{lim}}\frac{\mathrm{-3}+2\text{/}x}{\sqrt{4+5\text{/}{x}^2}}=\frac{\mathrm{-3}}{\sqrt{4}}=\frac{\mathrm{-3}}{2}.\hfill \end{array}$

Therefore, $f\left(x\right)$ approaches the horizontal asymptote $y=\frac{3}{2}$ as $x\to \infty$ and the horizontal asymptote $y=-\frac{3}{2}$ as $x\to \text{−}\infty$ as shown in the following graph.

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Determining end behavior for transcendental functions

The six basic trigonometric functions are periodic and do not approach a finite limit as $x\to \text{±}\infty .$ For example, $\text{sin}x$ oscillates between $1\text{and}\mathrm{-1}$ ( [link] ). The tangent function $x$ has an infinite number of vertical asymptotes as $x\to \text{±}\infty ;$ therefore, it does not approach a finite limit nor does it approach $\text{±}\infty$ as $x\to \text{±}\infty$ as shown in [link] .