Hello, synapsis!
Evaluating vertical asymptotes are easy for me to understand, just simply set the denominator to 0 and solve.
For the horizontal asymptote, it seems to be a comparison of the coefficients of the numerator and denominator.
That's the way it seems to be taught today, perhaps
without the reasoning behind it.
Example \(\displaystyle \;\)Find the horizontal asymptote (if any) for \(\displaystyle \L\,y \;=\;\frac{2x^2\,+\,5}{3x^2\,-\,4x}\)
We want to know what happens to the function as \(\displaystyle x\) gets very very large.
We see that, as \(\displaystyle x\to\infty\), the fraction goes to \(\displaystyle \,\frac{\infty}{\infty}\) . . . which is useless.
Here's the trick: \(\displaystyle \,\)Divide top and bottom by the highest power of \(\displaystyle x\)
in the denominator.
We divide top and bottom by \(\displaystyle x^2:\L\;\;\frac{\frac{2x^2}{x^2}\,+\,\frac{5}{x^2}}{\frac{3x^2}{x^2}\,-\,\frac{4x}{x^2}} \;=\;\frac{2\,+\,\frac{5}{x^2}}{3\,-\,\frac{4}{x}}\)
We can see that, as \(\displaystyle x\to\infty\), both \(\displaystyle \L\frac{5}{x^2}\) and \(\displaystyle \L\frac{4}{x}\) get smaller and smaller and go to 0.
So we have: \(\displaystyle \L\:y\;=\;\lim_{x\to\infty}\left(\frac{2\,+\,\frac{5}{x^2}}{3\,-\,\frac{4}{x}}\right)\;=\;\frac{2\,+\,0}{3\,-\,0}\;=\;\frac{2}{3}\)
Therefore, as \(\displaystyle x\) goes further and further to the right (and to the left)
\(\displaystyle \;\;\)the graph approches the horizontal line: \(\displaystyle \,y\:=\;\frac{2}{3}\)
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That was the case where the numerator and denominator had equal degrees.
What if the numerator is of higher degree?
Example \(\displaystyle \L\;y\;=\;\frac{3x^2\,+\,7x\,+\,5}{2x\,-\,3}\)
Divide top and bottom by \(\displaystyle x:\L\;\;\frac{\frac{3x^2}{x}\,+\,\frac{7x}{x}\,+\,\frac{5}{x}}{\frac{2x}{x}\,-\,\frac{3}{x}} \;= \;\frac{3x\,+\,7\,+\,\frac{5}{x}}{2\,-\,\frac{3}{x}}\)
We can see that, as \(\displaystyle x \to \infty\), both \(\displaystyle \L\frac{5}{x}\) and \(\displaystyle \L\frac{3}{x}\) go to zero.
\(\displaystyle \;\;\)but \(\displaystyle 3x\) gets larger and larger, of course.
So we have: \(\displaystyle \L\:y\;=\;\lim_{x\to\infty}\left(\frac{3x\,+\,7\,+\,\frac{5}{x}}{2\,-\,\frac{3}{x}}\right) \;= \;\frac{\infty\,+\,7\,+\,0}{2\,-\,0}\;=\;\infty\)
Hence, as \(\displaystyle x\) gets larger and larger, so does \(\displaystyle y\).
There is no horizontal asymptote.
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What if the numerator is of lesser degree?
Example \(\displaystyle \L\;y\;=\;\frac{2x\,-\,5}{3x^2\,+\,9}\)
Divide top and bottom by \(\displaystyle x:\L\;\;\frac{\frac{2x}{x}\,-\,\frac{5}{x}}{\frac{3x^2}{x} + \frac{9}{x}} \;= \;\frac{2\,-\,\frac{5}{x}}{3x\, +\, \frac{9}{x}}\)
We can see that, as \(\displaystyle x \to \infty\), both \(\displaystyle \L\frac{5}{x}\) and \(\displaystyle \L\frac{9}{x}\) go to 0
\(\displaystyle \;\;\)but \(\displaystyle 3x\) will continue to grow.
So we have: \(\displaystyle \L\,y\;=\;\lim_{x\to\infty}\left(\frac{2\,-\,\frac{5}{x}}{3x\,+\,\frac{9}{x}\right) \;= \;\frac{2\,-\,0}{\infty\,+\,0}\;=\;\frac{2}{\infty}\;=\;0\)
Therefore, the horizontal asymptote is \(\displaystyle y\,=\,0\), the x-axis.
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So if you ever forget those rules about exponents and leading coefficients,
you can make up examples like these and baby-talk your way through it.