$$f(x) =\frac{ (x - 1) }{(x^2 - 1)}$$ $$g(x) = \frac{1}{(x + 1)}$$
Is $f = g$? why?
Solution: answer is no. I don't get it. why can't it be same when it is the alt form?? Can anyone explain this further...
additional questions for further clarification: $$f(x) = \lim_{x\to 1}\frac{ (x^2 - 1) }{(x-1)} = infinite $$ $$f(x) = \lim_{x\to 1}\frac{ (x - 1)(x + 1) }{(x-1)} $$ $$f(x) = \lim_{x\to 1}(x+1)$$ $$ 1+1=2$$ How the function be justified that the limit can be apply to its alt form of a function which is not the same.
What they don't tell you until later in life is that a function is TWO things, a rule and a domain (plus the codomain).
That said, we often don't specify the domain since it is usually obvious. In the first function, for example, the domain can by any real number EXCEPT when the denominator is zero. Thus, the domain is the real line minus $1$ and $-1$.
The second function LOOKS the same, but the domain is the real number line minus $-1$.
It is a small difference, but since the domain is part of what defines the function, we say that the two functions are not the same.
However, we can of course write that $f=g$ for all $x\neq 1$.