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I have a function

$$f(x)=\frac{x^2-4}{x-2}$$

while solving this simply I get $f(1)=3,f(2)=\frac00,f(3)=5,\cdots$ the problem is when I solve this simply I get $$f(x)=x+2$$ and then I found except $f(2)$ all the values were same. Can someone help me where I did the mistake?

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    You might find [my answer here](http://math.stackexchange.com/a/2085545/276406) to be particularly helpful. In a nutshell, you've changed the *implicit* domain so you need to *explicitly* specify that the domain has not changed. – 2017-01-14

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You have to cancel $x-2$ then you have to give the condition$$f(x)=\frac{x^2-4}{x-2}$$ $$f(x)=\frac{(x-2)(x+2)}{(x-2)}$$ now you have to write $$f(x)=x+2\tag{where $x\ne2$}$$