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Why aren't these functions the same?by Whistlekins
Tags: functions 
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#1
Apr3013, 07:44 AM

P: 21

I have f(x) = (x^2+x2)/(x1) and g(x) = x+2
Now everyone would agree that f has a domain R\{1} and g has a domain R. Yet I can write (x^2+x2)/(x1) = x+2 So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this? 


#2
Apr3013, 07:50 AM

HW Helper
P: 3,516

[tex]\lim_{x\to 1}f(x) = 3[/tex] But just because the limit exists doesn't mean that the function is defined at that point. 


#3
Apr3013, 07:53 AM

P: 21

I understand that. But why can't I write g(x) = (x^2+x2)/(x1) = x+2 ?



#4
Apr3013, 07:54 AM

Sci Advisor
P: 821

Why aren't these functions the same?
##h : \mathbb{R} \to \mathbb{R}## with ##h(x) = \frac{x^2+x2}{x1}## and ##h(1) = 3## then that is indeed equal to g(x), but not equal to f(x). There is a difference between you can do something, and you did something. 


#5
Apr3013, 07:55 AM

Mentor
P: 18,099

[tex]\frac{x^2 + x  2}{x1} = x+2[/tex] is only valid for ##x\in \mathbb{R}## with ##x\neq 1##. For ##x=1##, it is not true. So we have that ##f(x) = x+2## for all ##x\in \mathbb{R}\setminus \{1\}##. The value ##f(1)## still isn't defined. That ##f(1)=3## somehow, is false. However, this is why limits are invented. So we can say that [tex]\lim_{x\rightarrow 1} f(x) = 3[/tex] So although ##f(1)## doesn't make sense, we can take the limit. The limit denotes the value that ##f(1)## would have been if it were defined in ##1## and if ##f## were to be continuous. 


#6
Apr3013, 08:02 AM

Sci Advisor
P: 821




#8
Apr3013, 08:09 AM

P: 21

So if I previously define the domain, I can't change that domain unless I write an entirely new function?
Would it be true that if h(x) = x+2 , for all x in R\{1}, then f = h? 


#10
Apr3013, 08:12 AM

P: 21

Cool, thanks for helping me clear my confusion. I guess that never really got explained to me by anyone and I never picked up on it.



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