What's up guys. I have a problem in my math class that I'm trying to figure out.

I've done the work here (looks better to the eye than typing it).

If anyone can explain to me why my professor does what he does, that'd be AWESOME.

https://imgur.com/a/rYLyM

I've done the work here (looks better to the eye than typing it).

If anyone can explain to me why my professor does what he does, that'd be AWESOME.

https://imgur.com/a/rYLyM

Just to clarify, I understand everything he's doing. But I'm wanting to understand WHY he's doing it.

Either your professor lying to you, or perhaps you're mistaken in the terminology between an "asymptote" and a simple "discontinuity".

Basically:

• An asymptote happens when, if you plug in a number, you get the result to be of the form {non-zero}/{zero}, e.g 3/0 in your example. You're dividing a non-zero by an incredibly small number, so it blows up to infinity.

• But, if you plug in and get 0/0 (your final """simplification"""), that is not an asymptote; 0/0 is indeterminant form, and can just mean that there is a simple hole or discontinuity at whatever number you plugged in.

In other words:

• (x+1)/(x-2) is your original equation. It has an asymptote at x = 2 because if you plug in x = 2, you get (3)/0, which is {non-zero}/{zero}. Okay. If you plug in x = 0, you simply get -1/2.

• Once you start factoring out an x and have and (x/x) in your equation, you create at a hole at x = 0, because now when you plug in 0 into (x^2 + x)/(x^2 - 2x), you get indeterminant form, (0 + 0)/(0 - 0) = 0/0.

**The limit is still exists and is -1/2, however.**

**tl;dr** after your professor manipulated the equation, he created a simple discontinuity or **hole** at x = 0, *not an asymptote*.

If you want visuals, and you don't have a graphing calculator on hand, you can use wolframalpha.com to plot things, because visuals are nice.

Edit: Changed wolfram alpha links to show the discontinuities

http://www.wolframalpha.com/input/?i=discontinuities+of+(x%5E2+%2B+x)%2F(x%5E2+-+2x)

http://www.wolframalpha.com/input/?i=discontinuities+of+(x+%2B+1)%2F(x+-+2)

https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/discontinuities-of-rational-functions/v/discontinuities-of-rational-functions

Basically:

• An asymptote happens when, if you plug in a number, you get the result to be of the form {non-zero}/{zero}, e.g 3/0 in your example. You're dividing a non-zero by an incredibly small number, so it blows up to infinity.

• But, if you plug in and get 0/0 (your final """simplification"""), that is not an asymptote; 0/0 is indeterminant form, and can just mean that there is a simple hole or discontinuity at whatever number you plugged in.

In other words:

• (x+1)/(x-2) is your original equation. It has an asymptote at x = 2 because if you plug in x = 2, you get (3)/0, which is {non-zero}/{zero}. Okay. If you plug in x = 0, you simply get -1/2.

• Once you start factoring out an x and have and (x/x) in your equation, you create at a hole at x = 0, because now when you plug in 0 into (x^2 + x)/(x^2 - 2x), you get indeterminant form, (0 + 0)/(0 - 0) = 0/0.

If you want visuals, and you don't have a graphing calculator on hand, you can use wolframalpha.com to plot things, because visuals are nice.

Edit: Changed wolfram alpha links to show the discontinuities

http://www.wolframalpha.com/input/?i=discontinuities+of+(x%5E2+%2B+x)%2F(x%5E2+-+2x)

http://www.wolframalpha.com/input/?i=discontinuities+of+(x+%2B+1)%2F(x+-+2)

https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/discontinuities-of-rational-functions/v/discontinuities-of-rational-functions

Last edited on

There is a single vertical asymptote at x=2.

If you multiply top and bottom of the fraction by x (not by the bizarre sequence of steps shown in the image) then you create a computational problem (division by 0) at x=0, but it is not an asymptote.

Either your teacher is having a bad day or what was intended was to start with that division of quadratics at the end and work back to what you have encircled, so removing any computational difficulty at x=0.

If you multiply top and bottom of the fraction by x (not by the bizarre sequence of steps shown in the image) then you create a computational problem (division by 0) at x=0, but it is not an asymptote.

Either your teacher is having a bad day or what was intended was to start with that division of quadratics at the end and work back to what you have encircled, so removing any computational difficulty at x=0.

Ganado,

Since my professor manipulated the equation, would

Since my professor manipulated the equation, would

`(x+1)/(x-2)`

still be the same as `x(x+1)/x(x-2)`

No, it is not *exactly* the same, but very close except at x = 0. That's still fundamentally the same problem, the numerator and denominator just aren't unfactored.

Multiplying by (x/x) doesn't create another asymptote, but it does create a simple discontinuity, otherwise known as a "hole" at x = 0, where the old equation was defined, but the modified equation is not.

Edit: It might be more formally correct to say "removable singularity", but I personally have never used that term in my schooling.

https://en.wikipedia.org/wiki/Classification_of_discontinuities

https://en.wikipedia.org/wiki/Removable_singularity

Multiplying by (x/x) doesn't create another asymptote, but it does create a simple discontinuity, otherwise known as a "hole" at x = 0, where the old equation was defined, but the modified equation is not.

Edit: It might be more formally correct to say "removable singularity", but I personally have never used that term in my schooling.

https://en.wikipedia.org/wiki/Classification_of_discontinuities

https://en.wikipedia.org/wiki/Removable_singularity

Last edited on

I’d like to point out that your professor is playing magical math with you. Kind of like those stupid MENSA puzzles that work by changing the meaning of a word mid-sentence or like those dumb order of operations Facebook puzzles or the 1+1=5 puzzles.

**Your professor ***changed the equation*.

And tells you that he didn’t.

Yes he did.

He does it twice.

• In step one, he pretends to be fair and factors out an arbitrary x/x, creating a “removable discontinuity” at x.

• In step two, he drops all pretense and just multiplies by x/x.

I can create a discontinuity at any C I wish by simply factoring or multiplying by my preferred (C-x).

I’m a big fan of 7.

http://www.wolframalpha.com/input/?i=discontinuities+of+((7-x)(x+%2B+1))%2F((7-x)(x+-+2))

Step 3 is just*what he wanted* to have. It is an obvious result of multiplying by x/x, which he cannot avoid showing explicitly. Had he left out step 1, everyone would have immediately noticed what he did and complained.

This is a common magicians trick. He introduced step 1 to distract everyone from what he did at step 2 so that he could get step 3: blatant, undisguised modification of the function.

Hope this helps.

And tells you that he didn’t.

Yes he did.

He does it twice.

• In step one, he pretends to be fair and factors out an arbitrary x/x, creating a “removable discontinuity” at x.

• In step two, he drops all pretense and just multiplies by x/x.

I can create a discontinuity at any C I wish by simply factoring or multiplying by my preferred (C-x).

I’m a big fan of 7.

http://www.wolframalpha.com/input/?i=discontinuities+of+((7-x)(x+%2B+1))%2F((7-x)(x+-+2))

Step 3 is just

This is a common magicians trick. He introduced step 1 to distract everyone from what he did at step 2 so that he could get step 3: blatant, undisguised modification of the function.

Hope this helps.

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