Checkfate
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I am trying to fully understand this example from a textbook I am reading:
http://img59.imageshack.us/img59/9237/continuityyn8.jpg
What I am not understanding is how they are proving it for [-1,1].. The way I see it is they proved that the function is continuous for all values in it's domain...
For example, I thought up this problem on my own to help me understand :
Given f(x)=1-\frac{1}{x-4}, prove that f(x) is continuous in the interval [-1,30] (Obviously it's not continuous at x=4.) The problem is that I don't see the connection between the interval and the solution...
I can just go ahead and prove that \lim_{x\rightarrow a}f(x)=f(a)... Which was stated in my text as meaning that the function is continuous... which it obviously isnt.
\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}(1-\frac{1}{x-4})
=1-\lim_{x\rightarrow a}\frac{1}{x-4}
=1-\frac{1}{a-4}
=f(a)
Can someone cure my confusion? Thanks guys.
http://img59.imageshack.us/img59/9237/continuityyn8.jpg
What I am not understanding is how they are proving it for [-1,1].. The way I see it is they proved that the function is continuous for all values in it's domain...
For example, I thought up this problem on my own to help me understand :
Given f(x)=1-\frac{1}{x-4}, prove that f(x) is continuous in the interval [-1,30] (Obviously it's not continuous at x=4.) The problem is that I don't see the connection between the interval and the solution...
I can just go ahead and prove that \lim_{x\rightarrow a}f(x)=f(a)... Which was stated in my text as meaning that the function is continuous... which it obviously isnt.
\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}(1-\frac{1}{x-4})
=1-\lim_{x\rightarrow a}\frac{1}{x-4}
=1-\frac{1}{a-4}
=f(a)
Can someone cure my confusion? Thanks guys.
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