• Why does $\lim\limits_{x\rightarrow a}(f-g)(x)=0$ then $f(x)\sim_{x\rightarrow a}g(x)$ false
  • and $f(x)\sim_{x\rightarrow a}g(x)$ implies $\lim\limits_{x\rightarrow a}(f-g)(x)=0$ false?

I was given the fact that $f(t)=t \ \& \ f(t)=t^2$ aren't equivalent aroud $0$ but have same limits, but aren't they equivalent as far as they have same limits!

Furthermore I was also given the fact that $f(t)=t \ \& \ g(t)=t+\ln t$ equivalent when $+\infty$ even if their difference is $+\infty$

  • $\begingroup$ What notion of equivalence are you using? $\endgroup$ – Travis Apr 23 '16 at 11:18
  • 1
    $\begingroup$ normally equivalence is not based on difference, but on quotient (at least in the current context of limits) and hence your implications are false in general. By definition $f$ is equivalent to $g$ if $f/g \to 1$. Note that this is not the same thing as $f - g \to 0$. If $g$ is bounded and away from $0$ then both $f - g \to 0$ and $f/g \to 1$ follow from each other. Thus in this particular case both your implications will be true. $\endgroup$ – Paramanand Singh Apr 23 '16 at 11:24
  • $\begingroup$ Does the answer to your previous quesiton help or do you have any questions ? Any reaction is appreciated. math.stackexchange.com/questions/1754323/… $\endgroup$ – callculus Apr 23 '16 at 12:56

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