Horizontal Asymptotes.

**habibixox** · November 26th, 2011, 11:31 AM

y = 19 x
(x4 +1)[1/4]
How can I find the horizontal asymptotes?

Can I apply L'Hopital's rule?

**mr fantastic** · November 26th, 2011, 11:34 AM

Originally Posted by habibixox

y = 19 x
(x4 +1)[1/4]
How can I find the horizontal asymptotes?

Can I apply L'Hopital's rule?

Your function is unintelligible. Please learn to use latex. Either that or use brackets, ^ for raising to a power etc.

**habibixox** · November 26th, 2011, 05:21 PM

[(19x)/((x^4 +1)^(1/4))]
sorry.

**mr fantastic** · November 26th, 2011, 09:09 PM

Originally Posted by habibixox

[(19x)/((x^4 +1)^(1/4))]
sorry.

To find a horizontal asymptote, you need to consider the limit $\lim_{x \to \pm \infty} \frac{19x}{(x^4 + 1)^{1/4}}$ .

It is not difficult to prove that the limit is equal to 19. Therefore ....

**habibixox** · November 27th, 2011, 08:01 AM

I can use L'Hopitals rule right? Because it would be inf/inf?

**habibixox** · November 27th, 2011, 08:04 AM

I just guessed the answers in my online hw and got 19 and -19 as the answers. But honestly I don't know how to do that... can someone show me the steps please?

**skeeter** · November 27th, 2011, 08:16 AM

$\lim_{x \to \pm \infty} \frac{19x}{\sqrt[4]{x^4+1}} =$

$\lim_{x \to \pm \infty} \frac{\frac{19x}{\sqrt[4]{x^4}}}{\sqrt[4]{\frac{x^4}{x^4}+\frac{1}{x^4}}} =$

$19 \lim_{x \to \pm \infty} \frac{x}{|x|\sqrt[4]{1+\frac{1}{x^4}}} =$

$-19$ as $x \to -\infty$

$19$ as $x \to \infty$

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