Vertical Asymptotes and squaring

[Glissando]

Homework Statement

Find the vertical asymptotes of:

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

Homework Equations

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-?

 

[tiny-tim]

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 …

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+ ? ​

 

[Glissando]

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

 

[HallsofIvy]

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.}