Answer:
I. and II. are correct
Step-by-step explanation:
The line y=0 is a horizontal asymptote in both directions, so ...
[tex]\lim\limits_{x\to -\infty}{f(x)}=\lim\limits_{x\to\infty}{f(x)}=0[/tex]
The function value f(0) is 0, so the limit is the same from either direction:
[tex]\lim\limits_{x\to 0^-}{f(x)}=\lim\limits_{x\to 0^+}{f(x)}=0[/tex]
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The expression for statement III can be simplified:
[tex]\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)}[/tex]
The limit of this is 0 from either direction, so the limit does exist at x=1.
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We consider that ...
both statements I. and II. are correct.
