# Vertical Asymptotes and squaring

• Glissando
In summary: There is no "vertical asymptote".In summary, when finding the vertical asymptotes of a function, one must first find the zeroes of the denominator and then find the limit as it approaches that number. When considering the limit as x approaches a number from the left or right, the notation "0+" or "0-" represents a number infinitesimally above or below that number, respectively. The bottom of the fraction, not the whole fraction, becomes 0+ or 0- when taking the limit. The square of a number just below (or above) the number being approached is just below (or above) the square of the number being approached. By subtracting the square of the number being approached from the bottom of
Glissando

## Homework Statement

Find the vertical asymptotes of:

f(x) = (x^2-1)/(x^2-4)

Limits, zeroes

## The Attempt at a Solution

Sorry about the mess...not too sure how to type some of this ):

lim (2^2-1)/(2^2-4) = 0+
x->2+

lim
x->2- (2^2-1)/(2(-)^2-4) = 0+

and lim as x -> -2...

I essentially understand how to do these problems (Find the zeroes of the denominator and find the limit as it approaches that number). What I don't understand is what happens to the (-) when x^2. For example in the above question, when the limit approaches 2 from the left (2-), when you square it does the (-) become positive, yielding 0+? Or does it not follow rules of squaring negative numbers and remain as 0-?

Hi Glissando!

(is that a reference to music, or to baseball? )

The real problem is that you're confusing yourself by writing it like this
Glissando said:
lim (22-1)/(22-4) = 0+
x->2+

lim
x->2- (22-1)/(2(-)2-4) = 0+

and lim as x -> -2...

(try using the X2 icon just above the Reply box )

first, if you use the "2+" (or "2-") notation, that's instead of using lim …

it means "a number infinitesimally above (or below) 2"​

second, it's only the bottom that becomes 0+ (or 0-), not the whole fraction, isn't it?

anyway, (2-)2 is the square of a number just below 2, so it's just below 4, ie (2-)2 = 4- …

(and (-2+)2 = 4- also, and (-2-)2 = 4+)

you then subtract 4 from 4- …

is that 0- or 0+ ?

tiny-tim said:
Hi Glissando!

(is that a reference to music, or to baseball? )

The real problem is that you're confusing yourself by writing it like this

(try using the X2 icon just above the Reply box )

first, if you use the "2+" (or "2-") notation, that's instead of using lim …

it means "a number infinitesimally above (or below) 2"​

second, it's only the bottom that becomes 0+ (or 0-), not the whole fraction, isn't it?

anyway, (2-)2 is the square of a number just below 2, so it's just below 4, ie (2-)2 = 4- …

(and (-2+)2 = 4- also, and (-2-)2 = 4+)

you then subtract 4 from 4- …

is that 0- or 0+ ?

Oh WOW that just made things a lot simpler. Thanks so much (: *it's music btw.

Thanks! <3

Using specific values, if x= 1.99< 2 then (1.99)2=3.9601 so that 1.992- 1= 2.9601 and 1.992- 4= -0.0399. (1.992- 1)/(1.992- 4)= -74.2 approximately. That is enough to tell you that the "limit", as x goes to 2 from below, is negative infinity. Similarly if x= 2.01> 2 then 2.012= 4.0401 so that 2.012- 1= 3.0401 and 2.012- 2= 0.0401. (2.012- 1)/(2.012- 4)= 3.0401/.0401= 75.8 approximately. That is enough to tell you that the "limit", as x goes to 2 from above, is positive infinity. And that means that the limit itself does not exist, even in the "infinite" sense.

## 1. What is a vertical asymptote?

A vertical asymptote is a vertical line on a graph where the function approaches infinity or negative infinity as the input value approaches a certain value.

## 2. How do you find the vertical asymptotes of a function?

To find the vertical asymptotes of a function, set the denominator of the function equal to zero and solve for the input values that make the denominator equal to zero. These input values are the vertical asymptotes.

## 3. What is the difference between a vertical asymptote and a horizontal asymptote?

A vertical asymptote is a vertical line on a graph where the function approaches infinity or negative infinity, while a horizontal asymptote is a horizontal line where the function approaches a constant value as the input values approach infinity or negative infinity.

## 4. How does squaring affect the vertical asymptote of a function?

Squaring a function can change the location of the vertical asymptotes. In some cases, it may eliminate or create new vertical asymptotes, while in other cases, it may leave the original vertical asymptotes unchanged.

## 5. Can a function have more than one vertical asymptote?

Yes, a function can have multiple vertical asymptotes. This can occur when there are multiple input values that make the denominator of the function equal to zero, creating multiple vertical lines on the graph where the function approaches infinity or negative infinity.

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