[SOLVED] asymptotes!

**fabxx** · October 17th 2008, 05:30 PM

Hi,
This is on the Barron SAT Math II page 17 # 28

Which of the following lines are asymptotes of the graph of y= $frac{x}{x+1}$ ?

I. x=1
II. x=-1
III. y=1

(For the answer you can choose two or more, or even none if its all incorrect.)

I know that x=1 is WRONG, and that x=-1 is CORRECT. But what about y=1? Why is it also an asymptote?

Thanks in advance

**Chris L T521** · October 17th 2008, 05:33 PM

Originally Posted by fabxx

Hi,
This is on the Barron SAT Math II page 17 # 28

Which of the following lines are asymptotes of the graph of y= $frac{x}{x+1}$ ?

I. x=1
II. x=-1
III. y=1

(For the answer you can choose two or more, or even none if its all incorrect.)

I know that x=1 is WRONG, Yup

and that x=-1 is CORRECT. Yup

But what about y=1? Why is it also an asymptote?

Thanks in advance

Examine the behavior of the function as $x\to{\pm\infty}$

So evaluate $\lim_{x\to\pm\infty}\frac{x}{x+1}$ and see what you get. This will be the horizontal asymptote with an equation of the form $y=k$ .

--Chris

**11rdc11** · October 17th 2008, 05:43 PM

Originally Posted by fabxx

Hi,
This is on the Barron SAT Math II page 17 # 28

Which of the following lines are asymptotes of the graph of y= $frac{x}{x+1}$ ?

I. x=1
II. x=-1
III. y=1

(For the answer you can choose two or more, or even none if its all incorrect.)

I know that x=1 is WRONG, and that x=-1 is CORRECT. But what about y=1? Why is it also an asymptote?

Thanks in advance

$\lim_{x \to \pm \infty}~ \frac{x}{x+1}$

You can solve this using L'hopital rule or divide by the largest degree of x

$\lim_{x \to \pm \infty}~ \frac{\frac{x}{x}}{\frac{x}{x} +\frac{1}{x}}$

which simplfies to

$\lim_{x \to \pm \infty}~ \frac{1}{1+\frac{1}{x}} = 1$

**Chris L T521** · October 17th 2008, 05:50 PM

Originally Posted by 11rdc11

$\lim_{x \to \pm \infty}~ \frac{x}{x+1}$

You can solve this using L'hopital rule or divide by the largest degree of x

$\lim_{x \to \pm \infty}~ \frac{\frac{x}{x}}{\frac{x}{x} +\frac{1}{x}}$

which simplfies to

$\lim_{x \to \pm \infty}~ \frac{1}{1+\frac{1}{x}} = 1$

Ha.

Or you can cheat...

$\lim_{x\to\infty}\frac{x}{x+1}=\frac{1}{1}=1$ .

In general,

$\lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\dots+b_2x^2+b_1x+b_0}=\left\{\begin{array}{rl} 0&\text{when }m>n\\ \displaystyle\frac{a_n}{b_m}&\text{when }m=n\\\infty&\text{when }m<n\end{array}\right.$

--Chris

**11rdc11** · October 17th 2008, 05:56 PM

Originally Posted by Chris L T521

Ha.

Or you can cheat...

$\lim_{x\to\infty}\frac{x}{x+1}=\frac{1}{1}=1$ .

In general,

$\lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\dots+b_2x^2+b_1x+b_0}=\left\{\begin{array}{rl} 0&\text{when }m>n\\ \displaystyle\frac{a_n}{b_m}&\text{when }m=n\\\infty&\text{when }m<n\end{array}\right.$

--Chris

Lol

**Jhevon** · October 17th 2008, 06:29 PM

Originally Posted by Chris L T521

Ha.

Or you can cheat...

$\lim_{x\to\infty}\frac{x}{x+1}=\frac{1}{1}=1$ .

In general,

$\lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\dots+b_2x^2+b_1x+b_0}=\left\{\begin{array}{rl} 0&\text{when }m>n\\ \displaystyle\frac{a_n}{b_m}&\text{when }m=n\\\infty&\text{when }m<n\end{array}\right.$

--Chris

excellent summary. the OP should bare in mind that he should watch his signs for the m < n case. you can get $- \infty$ depending on the coefficients of $x^n$ and $x^m$ .

a similar outcome is of course generated by taking the limit as $x \to - \infty$ , it this case, you also have to watch out how the values of n and m change the signs

**Chris L T521** · October 17th 2008, 06:30 PM

Originally Posted by Jhevon

excellent summary. the OP should bare in mind that he should watch his signs for the m < n case. you can get $- \infty$ depending on the coefficients of $x^n$ and $x^m$ .

a similar outcome is of course generated by taking the limit as $x \to - \infty$

Yup...I just focused on the case with $x\to\infty$

I'm to tired to type out what happens when $x\to-\infty$

--Chris

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