# Slant Asymptote

• Oct 23rd 2011, 03:05 PM
joaofonseca
Slant Asymptote
Given the function $\displaystyle f(x)=\frac{3x^4+4}{x^3-3x}$ I have found, by long division, that there is a slant asymptote, y=3x.
Further, I wanted to find the same result by the classic limit definition.

$\displaystyle \lim_{x \mapsto \infty}\left [ \frac{3x^4+4}{x^3-3x} \right ]$

$\displaystyle \lim_{x \mapsto \infty}\left [ \frac{x^4(1+\frac{4}{x^4})}{x^4(\frac{1}{x}-\frac{3}{x^3})} \right ]$

$\displaystyle \lim_{x \mapsto \infty}\left [ \frac{1+\frac{4}{x^4}}{\frac{1}{x}-\frac{3}{x^3}} \right ]=\frac{1+0}{0-0}$

I ended up in an indeterminate form.The result should had be the same!What's wrong?
• Oct 23rd 2011, 03:26 PM
skeeter
Re: Slant Asymptote
$\displaystyle \frac{1}{0}$ is not an indeterminate form.
• Oct 23rd 2011, 04:06 PM
joaofonseca
Re: Slant Asymptote
I made a mistake in the simplification.The final result is $\displaystyle \frac{3}{0}$. The problem is division by zero!
• Oct 23rd 2011, 04:08 PM
skeeter
Re: Slant Asymptote
Quote:

Originally Posted by joaofonseca
I made a mistake in the simplification.The final result is $\displaystyle \frac{3}{0}$. The problem is division by zero!

which means the limit does not exist.
• Oct 24th 2011, 02:50 AM
joaofonseca
Re: Slant Asymptote
The limit exist!Look at the graph:

Attachment 22670
• Oct 24th 2011, 03:04 AM
skeeter
Re: Slant Asymptote
I'm afraid you are confused ... the limit as $\displaystyle x \to \infty$ will determine a horizontal asymptote, but not a slant asymptote.

After you do the long division, you get an expression of the form $\displaystyle y = mx + b + r(x)$, where $\displaystyle r(x)$ is a remainder that tends to 0 as $\displaystyle x \to \infty$ , telling you that the function value approaches the line $\displaystyle y = mx+b$