Slant Asymptote

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October 23rd 2011, 03:05 PM
joaofonseca

Slant Asymptote

Given the function $f(x)=\frac{3x^4+4}{x^3-3x}$ 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.

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

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

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

I ended up in an indeterminate form.The result should had be the same!What's wrong?
October 23rd 2011, 03:26 PM
skeeter

Re: Slant Asymptote

$\frac{1}{0}$ is not an indeterminate form.
October 23rd 2011, 04:06 PM
joaofonseca

Re: Slant Asymptote

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

which means the limit does not exist.
October 24th 2011, 02:50 AM
joaofonseca

1 Attachment(s)

Re: Slant Asymptote

The limit exist!Look at the graph:

Attachment 22670
October 24th 2011, 03:04 AM
skeeter

Re: Slant Asymptote

I'm afraid you are confused ... the limit as $x \to \infty$ will determine a horizontal asymptote, but not a slant asymptote.

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