in the function f(x) = 1 + 1/x, when x= 0 does f(x) remain real? apparently not (seeing the graph), but why? shouldnt f(x) = 1? is my graph wrong?
It makes no sense to talk about the value of f(x) when x = 0.
Assuming you mean $\displaystyle f(x) = 1 + \frac{1}{x}$:
$\displaystyle \lim_{x \rightarrow 0^+} f(x) = + \infty$
$\displaystyle \lim_{x \rightarrow 0^-} f(x) = - \infty$
$\displaystyle \lim_{x \rightarrow + \infty} f(x) = 1$
$\displaystyle \lim_{x \rightarrow - \infty} f(x) = 1$
dom f = R\{0}. x = 0 is a vertical asymptote.
ran f = R\{1}. y = 1 is a horizontal asymptote.
Suppose $\displaystyle 1+\frac{1}{x}=1$
This gives $\displaystyle \frac{x+1}{x}=1 \Rightarrow x+1=x \Rightarrow 0=1$
Hence there does not exist an $\displaystyle x$ such that $\displaystyle 1+\frac{1}{x}=1$.
EDIT: this is quite apparent since for $\displaystyle 1+\frac{1}{x}=1 \Rightarrow \frac{1}{x}=0$ which cannot happen (since it leads to the same contradiction as before).