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Interpreting a function based on it's equation. 
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#19
Feb2012, 06:20 PM

P: 76




#20
Feb2012, 06:35 PM

Mentor
P: 21,409

Yes, the expression is one of several indeterminate forms, but the limit is 2, which is not indeterminate. In any case, a limit is never indeterminate.



#21
Feb2012, 07:44 PM

P: 76

My point is that the expression [itex]\frac{x^{2}1}{x1}[/itex] is not itself indeterminate. Evaluated at any realnumber value except [itex]x=1[/itex], it is singlevalued and welldefined. However, at [itex]x=1[/itex], its denominator is zero, so its value is undefined. The place where I am having some problem with terminology is here: If we consider the above expression as a rational function [itex]R(x)=\frac{P(x)}{Q(x)}[/itex] with [itex]P(x)=x^{2}1[/itex] and [itex]Q(x)=x1[/itex], we can use the heuristic I mentioned to understand why [itex]lim _{x \to 1}R(x)[/itex] yields, at least in my words, an "indeterminate form": I say that it is a heuristic because I am not letting [itex]P(x)[/itex] and [itex]Q(x)[/itex] vary simultaneously as they would when finding [itex]lim_{x \to 1}R(x)[/itex], rather I am considering either [itex]P(1)=0[/itex] or [itex]Q(1)=0[/itex] and allowing the other function of [itex]x[/itex] to vary as [itex]x \to 1[/itex] to show why, as I understand it, the form of the limit is indeterminate. Basically if I let, [itex]P(x)[/itex] vary, it yields a limit of the form [itex]lim_{x \to 1}=\frac{P(x)}{0}[/itex], which identically undefined everywhere except at [itex]x=1[/itex]; whereas, if I let [itex]Q(x)[/itex] vary, it yields a limit of the form [itex]lim_{x \to 0}=\frac{0}{Q(x)}[/itex], which identically 0 everywhere except at [itex]x=1[/itex]. Because the limit has a different value depending on whether [itex]P(x)[/itex] varies [itex]Q(x)[/itex], the value of the limit in the form given is indeterminate. The limit does exist and is equal to the value already mentioned, but it is impossible to tell what its value is. I'm not sure if I am expressing myself in the standard way, but that is how I understand the problem. I would appreciate any correction. 


#22
Feb2012, 07:53 PM

PF Gold
P: 1,957

[tex]\lim_{(x,y)\to(1,1)} \frac{y^2  1}{x1}[/tex] Which is true, but much less meaningful. The standard sense of the limit has the two parts of the fraction tending toward 1 simultaneously. It's much less useful and much less meaningful otherwise. 


#23
Feb2012, 08:17 PM

P: 76




#24
Feb2012, 08:37 PM

Mentor
P: 21,409

We can evaluate the limit to find that the discontinuity at x = 1 is removable, meaning that there is a "hole" at (1, 2). 


#25
Feb2012, 08:46 PM

Sci Advisor
P: 839

[itex]\frac{y^21}{x1}[/itex] is indeed in indeterminate as (x,y)>(1,1) because under the substitution (x,y)=(1,1) it becomes 0/0, and not because any dependence on paths. You need to understand "undefined" "intermediate" and "limit does not exist" are three different terms. Path dependence uses the third not the second. 


#26
Feb2112, 12:26 AM

P: 115

Undefined  a point for an equation does not exist. Found by dividing by zero, squaring a negative (other ways??) Indeterminate  when a form is indeterminate, it takes the form of 0/0, infin/infin, infin  infin, 0 x infin, infin ^ 0, 0 ^ 0, c ^ infin (c is some constant) Lim does not exist  right hand side and left hand side limits are not equal that right? 


#27
Feb2112, 02:15 PM

P: 76




#28
Feb2112, 03:59 PM

Mentor
P: 21,409

Possible ways an expression can fail to exist: division by zero; taking the square root (or an even root) of a negative number  there's no problem with squaring a negative number; attempting to evaluate a function at a value outside its domain (e.g., ln(0)), and so on. [tex]\lim_{x \to 0} \sqrt{x} \text{ does not exist}[/tex] 


#29
Sep2012, 09:13 PM

P: 115

Wow it's almost a year since I've first seen this material and I am running back into it. Geeze I remember vividly how much this made my head spin :D lol  it was one of the most confusing things for me in calculus along with l'hospitals rule etc. Now it so much more clear reading the textbook and understanding it all. And I remember that this thread provided lot's of help too at the time and I made reference to it a few times afterwards. So thanks a lot everyone. All the responses are top notch!



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