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Math Help - real function?

  1. #1
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    Question 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?
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  2. #2
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    Quote Originally Posted by the kopite View Post
    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.
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  3. #3
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    which means that f(x)= 1 + (1/x) cannot be real for x = 0? that was the question.
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  4. #4
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    Quote Originally Posted by the kopite View Post
    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.
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  5. #5
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    ?
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  6. #6
    Super Member Showcase_22's Avatar
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    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).
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  7. #7
    Junior Member Nerdfighter's Avatar
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    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.
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