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?

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- May 31st 2009, 10:53 PMthe kopitereal 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 PMmr fantastic
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. - Jun 1st 2009, 12:30 AMthe kopite
which means that f(x)= 1 + (1/x) cannot be real for x = 0? that was the question.

- Jun 1st 2009, 12:32 AMmr fantastic
- Jun 1st 2009, 12:39 AMthe kopite
?

- Jun 1st 2009, 05:15 AMShowcase_22
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). - Jun 1st 2009, 04:46 PMNerdfighter
Basically, you can never divide by zero. $\displaystyle \frac{1}{0}$ is undefined. It simply does not exist, and there is nothing wrong with your graph.