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Harman [31]
3 years ago
11

Thanks for the help!​

Mathematics
1 answer:
Serhud [2]3 years ago
6 0

Answer:

  I. and II. are correct

Step-by-step explanation:

The line y=0 is a horizontal asymptote in both directions, so ...

  \lim\limits_{x\to -\infty}{f(x)}=\lim\limits_{x\to\infty}{f(x)}=0

The function value f(0) is 0, so the limit is the same from either direction:

  \lim\limits_{x\to 0^-}{f(x)}=\lim\limits_{x\to 0^+}{f(x)}=0

__

The expression for statement III can be simplified:

  \dfrac{f(x)-\dfrac{1}{2}}{x-1}=\dfrac{\dfrac{x}{1+x^2}-\dfrac{1}{2}}{x-1}=\dfrac{2x-(1+x^2)}{2(1+x^2)(x-1)}=\dfrac{-(x-1)^2}{2(x-1)(x^2+1)}\\\\=\dfrac{1-x}{2(x^2+1)}

The limit of this is 0 from either direction, so the limit does exist at x=1.

_____

We consider that ...

  both statements I. and II. are correct.

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Answer:

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Step-by-step explanation:

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