As you age, you grow taller—but not without bound!
Eventually, each person settles at a particular height.
Turn off an oven and open up the door—the oven cools down.
But it doesn't keep getting colder and colder!
Eventually, it settles at room temperature.
Tie an object to the end of a spring, and give it a tug.
It bounces ^{up} and _{down}, ^{up} and _{down}, ^{up} and _{down}.
Eventually, though, it settles at a particular height.
If you model these reallife scenarios using mathematics, then you'll find yourself dealing with horizontal asymptotes!
As discussed in Introduction to Asymptotes,
an asymptote is a
curve (usually a line)
Here's how this general definition is ‘specialized’ to get a
horizontal asymptote:
that another curve gets arbitrarily close to as $\,x\,$ approaches $\,+\infty\,$ or $\,\infty\,$.
What's the key to a horizontal asymptote situation? as inputs get arbitrarily large (big and positive, or big and negative).
A situation where something ‘levels out’ or reaches a steadystate
This section gives a precise discussion of horizontal asymptotes, 
The red line is a horizontal asymptote for the blue curve. As $\,x\rightarrow\pm\infty\,$, the blue curve approaches the red line. That is, as $\,x\rightarrow\pm\infty\,$, the blue curve outputs approach a specific finite number. 









Let $\displaystyle\,R(x) = \frac{3x1}{5x^2 + 2x  1}\,$.
To examine $\,R\,$ for horizontal asymptote behavior, you must understand what the outputs
look like when the inputs are very big.
More precisely, you must understand what the outputs look like as $\,x\rightarrow\pm\infty\,$.
Let's investigate what happens when $\,x\,$ is big and positive.
As $\,x\rightarrow\infty\,$, the numerator goes to infinity. (Plug big positive numbers into $\,3x  1\,$.)
As $\,x\rightarrow\infty\,$, the denominator goes to infinity. (Plug big positive numbers into $\,5x^2 + 2x  1\,$.)
So, as $\,x\rightarrow\infty\,$, we have something like looks like $\,\frac{\text{big positive}}{\text{big positive}}\,$.
How useful is this?
Not very useful at all!
Let's see if we can rename the function to better understand what happens when $\,x\,$ is big:
The highest power that appears on any $\,x\,$ is $\,\color{red}{2}\,$.
So, the technique we'll use is to multiply both numerator and denominator by $\displaystyle\frac{1}{x^{\color{red}{2}}}\,$.
Notice that we're not changing any output values, since we're just multiplying by one!
$$
\cssId{s66}{R(x)}
\cssId{s67}{= \frac{3x1}{5x^2 + 2x  1}} \ \
\cssId{s68}{=\ \ \frac{(3x1)}{(5x^2 + 2x  1)}\cdot\frac{\frac{1}{x^2}}{\frac{1}{x^2}}} \ \
\cssId{s69}{=\ \ \frac{\frac 3x  \frac 1{x^2}}{5 + \frac{2}{x}  \frac{1}{x^2}}}
$$
At first glance, it may look like our renaming has made things more complicated.
However, for the purpose of understanding what the outputs look like when $\,x\,$ is big,
this is a much better name!
Why?
As $\,x\rightarrow\infty\,$ (or as $\,x\rightarrow \infty\,$),
most everything in sight tends to zero!
$$
\cssId{s74}{\frac{3}{x}\rightarrow 0,} \quad
\cssId{s75}{\frac{1}{x^2}\rightarrow 0,} \quad
\cssId{s76}{\frac{2}{x}\rightarrow 0}
$$
Therefore, as $\,x\,$ tends to $\,\pm\infty\,$,
$$
\cssId{s78}{\frac{\frac 3x  \frac 1{x^2}}{5 + \frac{2}{x}  \frac{1}{x^2}}}
\cssId{s79}{\rightarrow \frac{0  0}{5 + 0  0}}
\cssId{s80}{= \frac{0}{5}}
\cssId{s81}{= 0}
$$
So now we know!
When $\,x\,$ is big, the outputs are close to zero.
Thus, ‘$\,y = 0\,$’ is a horizontal asymptote.
Here's an easier (but less precise) way to think about what is happening:
Let $\displaystyle\,R(x) = \frac{3x1}{5x + 2}\,$.
Again, as $\,x\rightarrow \infty\,$, the fraction looks like $\,\frac{\text{big positive}}{\text{big positive}}\,$.
Not very useful in this form!
The highest power that appears on any $\,x\,$ is $\,\color{red}{1}\,$;
the technique to use is to multiply both numerator and denominator by $\displaystyle\frac{1}{x^{\color{red}{1}}} = \frac 1x\,$.
No output values are being changed in the following renaming, since we're just multiplying by one!
$$
\cssId{s101}{R(x)}
\cssId{s102}{= \frac{3x1}{5x + 2}} \ \
\cssId{s103}{=\ \ \frac{(3x1)}{(5x + 2)}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}} \ \
\cssId{s104}{=\ \ \frac{3  \frac 1{x}}{5 + \frac{2}{x}}}
$$
As $\,x\rightarrow\pm\infty\,$, both $\,\frac{1}{x}\,$ and $\,\frac{2}{x}\,$ tend to zero. Therefore, as $\,x\,$ tends to $\,\pm\infty\,$, $$ \cssId{s107}{\frac{3  \frac 1{x}}{5 + \frac{2}{x}}} \cssId{s108}{\rightarrow \frac{3  0}{5 + 0} = \frac{3}{5}} $$
So now we know!
When $\,x\,$ is big, the outputs are close to the finite number $\,\frac{3}{5}\,$.
Thus, ‘$\,y = \frac{3}{5}\,$’ is a horizontal asymptote.
Note that $\,\frac{3}{5}\,$ is the ratio of the leading coefficients of the numerator and denominator:
$\displaystyle\,R(x) = \frac{\boldsymbol{3}x1}{\boldsymbol{5}x + 2}$
Here's an easier (but less precise) way to think about what is happening:
Here's yet another easy (but less precise) way to think about what is happening:
In summary, we have:
NOTE: If $\deg N(x) > \deg D(x)$, then there is no horizontal asymptote.
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
