Defining a Triangle: Investigating $\lim_{x\rightarrow 1} \frac{y}{1-x}$

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SUMMARY

The discussion centers on the limit calculation of the expression $\lim_{x\rightarrow 1} \frac{y}{1-x}$ under the constraints defining a triangle: \(0 \leq y \leq 1\) and \(0 \leq x \leq 1-y\). It is established that the limit's behavior is contingent on the value of \(y\); if \(y\) is fixed between 0 and 1, \(x\) cannot approach 1 without violating the triangle's constraints. The conclusion reached is that if \(y = 1\), the limit does not exist, and the original problem was ultimately resolved through an alternative approach that avoided singularities.

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daudaudaudau
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If I define a triangle like this

[tex]0\le y\le 1[/tex] and [tex]0\le x\le 1-y[/tex],

then what is

[tex]\lim_{x\rightarrow 1} \frac{y}{1-x}[/tex] ?

As far as I can see, there are many possibilities, depending on how you "approach" [tex]x=1[/tex].
 
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What you are writing doesn't make sense. Since you are taking the limit only as x goes to 1, y is some fixed number between 0 and 1. If [itex]x\le 1-y[/itex] and y is not 1 then x cannot "approach" 1. If y= 1, then, of course, the limit does not exist.
 
Hmm, yeah, maybe it does not make much sense. Fortunately, I was able to fix my problem in another way, which gave a result with no singularities. So I don't have to solve this problem :) Phew.
 

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