evaluating horizontal asymptote

synapsis

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Jan 19, 2006
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evaluating veritical 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, and my book is vague on how this is done.

From my notes in class, it looks like if the leading coeff's are equal, i.e 3x^2/7x^2
the VA is 3/7....

but what if the leading coeff in the numerator is > or < the leading coeff in the denom....????

thanks for any help
 
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.
 
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