Asymptotes...help please

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December 12th 2007, 07:07 AM
Raleigh

Asymptotes...help please

I am unsure of where to start on these problems...I remember learning asymptotes, but I am a little confused by these problems....they seem different than what we learned...if someone could please help explain them to me, I would greatly appreciate it.

http://i103.photobucket.com/albums/m...1919/111_4.jpg
December 12th 2007, 07:41 AM
red_dog

16) $\lim_{x\to\pm\infty}\frac{3}{x-9}=0\Rightarrow y=0$ horizontal asymptote.

17) $f(x)=\frac{5x+7}{x^2+3x-10}=\frac{5x+7}{(x-2)(x+5)}$
$\lim_{x\nearrow 2}f(x)=-\infty, \ \lim_{x\searrow 2}f(x)=\infty\Rightarrow x=2$ vertical asymptote.

Find the limits in $x=-5$ and you'll get another vertical asymptote.
December 12th 2007, 07:50 AM
Raleigh

Thank you very much for your help! but I don't quite remember doing this...I am still a little bit confused about what you did...
December 12th 2007, 10:53 AM
earboth

Quote:

Originally Posted by Raleigh

Thank you very much for your help! but I don't quite remember doing this...I am still a little bit confused about what you did...

Hello,

a few general remarks:

an asymptote is a straight line or a curve which the graph of the function will approach when the |x|-values increases unlimited.
With your examples you only have to calculate the limit of the values of the function if x approaches infinity.

I assume that you know that $\lim_{x \mapsto \infty}\left(\frac1x \right) = 0$ and therefore

$\lim_{x \mapsto \infty}\left(\frac ax \right) = \lim_{x \mapsto \infty}\left(a \cdot \frac1x \right) = a \cdot \lim_{x \mapsto \infty}\left(\frac1x \right) = 0$ and $\lim_{x \mapsto \infty}\left(\frac1{x^n} \right) = 0~,~n \in \mathbb{N}$

Using this property the calculation of limits becomes easy:

$\lim_{x \mapsto \infty}\left(\frac{5x+7}{x^2+3x-10} \right)=\lim_{x \mapsto \infty}\left( \dfrac{x^2\left(\frac5x+\frac7{x^2}\right)} {x^2\left(1+\frac3x-\frac{10}{x^2}\right)} \right) = \frac01 = 0$

That means the asymptote is here: $y = 0$