Slant asymptote

[Math_Junkie]

Find the equation of the slant asymptote for the following function:

I'm reading ahead for my calculus class and I came across this question. I know that long division is somehow involved, but I don't understand why and what I'm doing.

Can anyone provide explanations how I can go about solving this?

 

[wjm11]

Find the equation of the slant asymptote...

I'm reading ahead for my calculus class and I came across this question. I know that long division is somehow involved

If you do the long division, you get y = 2x – 9 + ((46x + 4)/(x^2 + 5x)). As x becomes large (approaches infinity or negative infinity), the remainder (the ratio) approaches zero, so you can just forget about it. You’re left with

y = 2x – 9

This is the oblique (or slant) asymptote that the function approaches. Make sense?

This situation occurs when the degree of the numerator is one larger than the degree of the denominator.

 

[soroban]

Hello, Math_Junkie!

If you're familiar with limits, there's another (faster) approach.

However, I get a different answer.
I've come across this dilemma before and have no explanation for it.
Can anyone clear it up?

\(\displaystyle \text{Find the equation of the slant asymptote for: }\;y \:=\:\frac{2x^3 + x^2 + x + 4}{x^2+5x}\)

We want to know what the graph looks like when \(\displaystyle x\) gets very large.
So we take the limit of \(\displaystyle y\) as \(\displaystyle x \to \infty.\)


\(\displaystyle \text{We have: }\;\lim_{x\to\infty}y \;=\;\lim_{x\to\infty}\frac{2x^3 + x^2 + x + 4}{x^2 + 5x}\)


\(\displaystyle \text{Divide top and bottom by }x^2\text{, the highest power in the }denominator.\)

. . \(\displaystyle \lim_{x\to\infty}\frac{\dfrac{2x^3}{x^2} + \dfrac{x^2}{x^2} + \dfrac{x}{x^2} + \dfrac{4}{x^2}}{\dfrac{x^2}{x^2} + \dfrac{5x}{x^2}} \;=\; \lim_{x\to\infty}}\frac{2x + 1 + \dfrac{1}{x} + \dfrac{4}{x^2}}{1 + \dfrac{5}{x}} \;=\;\frac{2x+1 + 0 + 0}{1 + 0} \;=\;2x+1\)


\(\displaystyle \text{Therefore, the slant asymptote is: }\:y \:=\:2x+1\)

 

[lookagain]

However, I get a different answer.
I've come across this dilemma before and have no explanation for it.
Can anyone clear it up?

\(\displaystyle \text{Find the equation of the slant asymptote for: }\;y \:=\:\frac{2x^3 + x^2 + x + 4}{x^2+5x}\)

We want to know what the graph looks like when \(\displaystyle x\) gets very large.
So we take the limit of \(\displaystyle y\) as \(\displaystyle x \to \infty.\)


\(\displaystyle \text{We have: }\;\lim_{x\to\infty}y \;=\;\lim_{x\to\infty}\frac{2x^3 + x^2 + x + 4}{x^2 + 5x}\)


\(\displaystyle \text{Divide top and bottom by }x^2\text{, the highest power in the }denominator.\)

. . \(\displaystyle \lim_{x\to\infty}}\frac{2x + 1 + \dfrac{1}{x} + \dfrac{4}{x^2}}{1 + \dfrac{5}{x}} \;\) . . . This is not true.\(\displaystyle =\;\frac{2x+1 + 0 + 0}{1 + 0} \;=\;2x+1\)

The limit as \(\displaystyle x \to \infty\) for the expression on the far left is \(\displaystyle +\infty\).
Here, you would be showing that the function value is increasing without bound,
but it does not get you the the expression used in the slant asymptote.

\(\displaystyle \text{wjm11 has the correct equation of the asymptote.}\)