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Understanding Horizontal Asymptotes

confusedatmath

New member
Jan 2, 2014
14
I'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?

Screen Shot 2014-01-05 at 10.55.15 am.png
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What happens to a fraction when the denominator gets larger and larger without bound?
 

confusedatmath

New member
Jan 2, 2014
14
the fraction gets smaller, but dont we end up adding that to the value of c??


well if a/(x-b)^2 +c

let a=1 b=1 c=2

so we get

1/(x-1)^2 +2

lets make x =2

we get

1/1 +2 .... thats 3.... wait am i doing something wrong?

- - - Updated - - -

oh wait is because in the above a is negative.. so

it would be -1+2 < 2 ....
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The fraction goes to zero as the denominator goes to infinity, and so that's why the horizontal asymptote is $y=c$.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Let's make this really interesting by choosing some suitably large values for $x$.

Let's see what happens at $x = 1,000,000$, when $a = b = -1,c = 2$.

Then $f(x) = \dfrac{-1}{(999,999)^2} + 2$

$= \dfrac{-1}{999,998,000,001} + 2$

$\sim 1.999999999998999998$

At $x = -1,000,000$, we get:

$f(x) = \dfrac{-1}{(-1,000,001)^2} + 2$

$= \dfrac{-1}{1,000,002,000,001} + 2$

$\sim 1.999999999999000002$

Both of these numbers are really close to 2, right?

In general, we see that:

$\displaystyle \lim_{x \to \infty} \left(\frac{a}{(x + b)^2} + c\right)$

$\displaystyle = \lim_{x \to \infty} \frac{a}{(x + b)^2} + \lim_{x \to \infty} c$

$\displaystyle = \lim_{x \to \infty} \frac{a}{(x+b)^2} + c$

$\displaystyle = (a)\left(\lim_{x \to \infty}\frac{1}{x+b}\right)^2 + c$

$\displaystyle = (a)(0)^2 + c = 0 + c = c$

Similar reasoning holds to show that:

$\displaystyle \lim_{x \to -\infty} \left(\frac{a}{(x + b)^2} + c\right) = c$

as well.

(If you haven't been formally introduced to limits yet, all you need to know for this is the following (which hold under "suitably nice conditions" which are the case here):

1) the limit of a sum is the sum of the limits of each term in the sum
2) the limit of a product is the product of the limits of each factor in the product
3) if M(x) is a function that gets "infinitely big" as x does, then:

$$\lim_{x \to \infty} \frac{1}{M(x)} = 0$$

-by "get infinitely big as x does" I mean that for ANY positive integer $K$, there is always some positive integer $N$, so that if we have $x > N$, then $|M(x)| > K$ (typically, the integer $N$ will depend on $K$, bigger $K$'s usually need bigger $N$'s).).
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
I'm stuck at understanding why +c is the horizontal asymptote. Can someone please explain this? I get that the vertical asymptote is relating to (x+b), because the denominator cannot=0. But why does this kind of graph have a horizontal asymptote to begin with?

View attachment 1838
The short answer, if [tex]\displaystyle \begin{align*} y = c \end{align*}[/tex] then you end up with [tex]\displaystyle \begin{align*} 0 = \frac{a}{(x + b)^2} \end{align*}[/tex]. Is that possible to solve for x?

The long answer, consider the inverse relation. You should find that c becomes a vertical asymptote.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
the fraction gets smaller, but dont we end up adding that to the value of c??


well if a/(x-b)^2 +c

let a=1 b=1 c=2

so we get

1/(x-1)^2 +2

lets make x =2

we get

1/1 +2 .... thats 3.... wait am i doing something wrong?

- - - Updated - - -

oh wait is because in the above a is negative.. so

it would be -1+2 < 2 ....
The question was "what happens when x is large?" "2" is NOT large!

If x= 1000000, then 1/(1+ 1000000)+3= 3.000009999900000999990000099999 and for