1. ## [SOLVED] asymptotes!

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

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

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?

2. 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=$\displaystyle frac{x}{x+1}$?

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?

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

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

--Chris

3. 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=$\displaystyle frac{x}{x+1}$?

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?

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

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

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

which simplfies to

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

4. Originally Posted by 11rdc11
$\displaystyle \lim_{x \to \pm \infty}~ \frac{x}{x+1}$

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

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

which simplfies to

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

Or you can cheat...

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

In general,

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

--Chris

5. Originally Posted by Chris L T521
Ha.

Or you can cheat...

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

In general,

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

--Chris
Lol

6. Originally Posted by Chris L T521
Ha.

Or you can cheat...

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

In general,

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

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

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

7. 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 $\displaystyle - \infty$ depending on the coefficients of $\displaystyle x^n$ and $\displaystyle x^m$.

a similar outcome is of course generated by taking the limit as $\displaystyle x \to - \infty$
Yup...I just focused on the case with $\displaystyle x\to\infty$

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

--Chris