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Here's what we tried:

For every $\epsilon > 0$ there is a large number $K$ such that $|f(x)| < \epsilon$ when $x>K$.

Knowing that $K$ is a large positive number, take the positive absolute value of $f(x)$:

$$\displaystyle \frac{x}{1+x^2} < \epsilon$$

Solve for

$$x > \displaystyle \sqrt{\frac{x - \epsilon}{\epsilon}}$$

And thus $$K = \displaystyle \sqrt{\frac{x - \epsilon}{\epsilon}}$$.

Is it acceptable to have $K$ in terms of $x$?

  • 3
    We are not interested in finding the *cheapest* $K$ such that $x/(1+x^2) <\epsilon$ for $x>K$. Any $K$ that works will do. It is useful to have an eye for a simple expression $E(x)$ which goes to $0$ and such that $x/(1+x^2), at least for large $x$. Here it is easy, since for positive $x$, $x/(1+x^2). So if we make $1/x<\epsilon$, we will automatically have $x/(1+x^2) <\epsilon$.2011-10-14
  • 3
    No, it is certainly not acceptable to have $K$ in terms of $x$.2011-10-14

3 Answers 3