# real function?

• May 31st 2009, 10:53 PM
the kopite
real function?
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?
• May 31st 2009, 11:57 PM
mr fantastic
Quote:

Originally Posted by the kopite
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 $f(x) = 1 + \frac{1}{x}$:

$\lim_{x \rightarrow 0^+} f(x) = + \infty$

$\lim_{x \rightarrow 0^-} f(x) = - \infty$

$\lim_{x \rightarrow + \infty} f(x) = 1$

$\lim_{x \rightarrow - \infty} f(x) = 1$

dom f = R\{0}. x = 0 is a vertical asymptote.

ran f = R\{1}. y = 1 is a horizontal asymptote.
• Jun 1st 2009, 12:30 AM
the kopite
which means that f(x)= 1 + (1/x) cannot be real for x = 0? that was the question.
• Jun 1st 2009, 12:32 AM
mr fantastic
Quote:

Originally Posted by the kopite
which means that f(x)= 1 + (1/x) cannot be real for x = 0? that was the question.

I told you that it makes no sense to consider the value when x = 0. It is neither real nor not real. It is not defined.
• Jun 1st 2009, 12:39 AM
the kopite
?
• Jun 1st 2009, 05:15 AM
Showcase_22
Suppose $1+\frac{1}{x}=1$

This gives $\frac{x+1}{x}=1 \Rightarrow x+1=x \Rightarrow 0=1$

Hence there does not exist an $x$ such that $1+\frac{1}{x}=1$.

EDIT: this is quite apparent since for $1+\frac{1}{x}=1 \Rightarrow \frac{1}{x}=0$ which cannot happen (since it leads to the same contradiction as before).
• Jun 1st 2009, 04:46 PM
Nerdfighter
Basically, you can never divide by zero. $\frac{1}{0}$ is undefined. It simply does not exist, and there is nothing wrong with your graph.