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stellapi-lovesmath:

The PROPER definition of limits from M. Spivak’s Calc. book (the greatest of the Calc books)!
If you don’t use delta and epsilon in your limits, you aren’t actually doing limits.

stellapi-lovesmath:

The PROPER definition of limits from M. Spivak’s Calc. book (the greatest of the Calc books)!

If you don’t use delta and epsilon in your limits, you aren’t actually doing limits.

(Source: stelly-themathcat)

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Mean Girls Limit Problem

Mean Girls isn’t really a “new” movie, and I was like in grade 9 or 8 when it came out. And was amazed at the complex math that I didn’t understand in it. But after all this time, and me learning so much about math and being quite good with limits I thought that it was probably a really easy question so I googled it to check. However, I was quite interested with this problem, shown here:

Okay, so obviously the first thing I noticed was that 1 - cos^2 x = sin^2 x, that just might simplify things, but not very useful. 

Another easy thing to notice is that if we plug in “0” for x, we get 0/0 because sin 0 = 0, cos 0 = 1, and ln 1, meaning to what power we have to raise e to to get 1, being 0.

Normally people would just normally jump straight into L’Hopital’s rule which can be summarized into: 

If  exists and becomes 0/0 or infinity/infinity then  is correct to use.

If we follow our initial gut feeling to use L’Hopital’s rule the problem becomes :

lim [x->0] (-1/(1-x) - cosx)/(-2cosxsinx)  = -2/0 = inf. We can thus say that the limit is infinity or in some textbooks, doesn’t exist. Normally both would be right, however, we made a mistake here. The restriction of the first condition doesn’t come up often, but does here. 
A limit only exist if  . Which means that the limit of the number approaching from the negative and positive sides are equal. And in this case, 

lim_(x->0^-) (log(1-x)-sin(x))\/(1-cos^2(x)) = infinity\nlim_(x->0^+) (log(1-x)-sin(x))\/(1-cos^2(x)) = -infinity

Log being another way of writing ln, or natural logarithm, depending on what country you’re in, different notation is used. The reason for this difference in side limits is because of the ln x function and sin x function respectively: 

 

with the limit of approaching 0 slightly from the negative side, the ln(1-x) becomes a slightly positive 0 while the sin x becomes slightly negative and since it is being subtracted the entire top becomes a slightly positive 0, but with the limit approaching 0 slightly from the positive side, ln(1-x) becomes slightly negative and sin x becomes slightly positive and then is subtracted from the slightly negative ln(1-x). Thus the limit does not exist. 
This question requires a little be of a different approach than most limits, so in that sense, is a really good question. 

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14-billion-years-later:

warholsoup:

chicaintcheap:

sagebrown:

Rhythm 0, 1974
To test the limits of the relationship between performer and  audience, Abramović developed one of her most challenging (and  best-known) performances. She assigned a passive role to herself, with  the public being the force which would act on her.
Abramović had placed upon a table 72 objects that people were allowed  to use (a sign informed them) in any way that they chose. Some of these  were objects that could give pleasure, while others could be wielded to  inflict pain, or to harm her. Among them were scissors, a knife, a  whip, and, most notoriously, a gun and a single bullet. For six hours  the artist allowed the audience members to manipulate her body and  actions.
Initially, members of the audience reacted with caution and modesty,  but as time passed (and the artist remained impassive) several people  began to act quite aggressively. As Abramović described it later:
“The experience I learned was that…if you leave decision to the  public, you can be killed.” … “I felt really violated: they cut my  clothes, stuck rose thorns in my stomach, one person aimed the gun at my  head, and another took it away. It created an aggressive atmosphere.  After exactly 6 hours, as planned, I stood up and started walking toward  the public. Everyone ran away, escaping an actual confrontation.”  

14-billion-years-later:

warholsoup:

chicaintcheap:

sagebrown:

Rhythm 0, 1974

To test the limits of the relationship between performer and audience, Abramović developed one of her most challenging (and best-known) performances. She assigned a passive role to herself, with the public being the force which would act on her.

Abramović had placed upon a table 72 objects that people were allowed to use (a sign informed them) in any way that they chose. Some of these were objects that could give pleasure, while others could be wielded to inflict pain, or to harm her. Among them were scissors, a knife, a whip, and, most notoriously, a gun and a single bullet. For six hours the artist allowed the audience members to manipulate her body and actions.

Initially, members of the audience reacted with caution and modesty, but as time passed (and the artist remained impassive) several people began to act quite aggressively. As Abramović described it later:

“The experience I learned was that…if you leave decision to the public, you can be killed.” … “I felt really violated: they cut my clothes, stuck rose thorns in my stomach, one person aimed the gun at my head, and another took it away. It created an aggressive atmosphere. After exactly 6 hours, as planned, I stood up and started walking toward the public. Everyone ran away, escaping an actual confrontation.