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Could any one please elaborate how $\lim_{x \to \infty} f(x)/x=1$ can be derived from $1/\lambda<f(x)/x<\lambda$? I have taken it granted that $1/\lambda<f(x)/x<\lambda$.

The question is taken from below proof- enter image description here

PS: I made a mistake in previous post so, I am re-posting.

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    $\begingroup$ "These inequalities hold for all $\lambda > 1$"... $\endgroup$
    – amsmath
    Oct 24, 2019 at 3:49
  • $\begingroup$ @amsmath not wise enough to get through that particular line :) $\endgroup$ Oct 24, 2019 at 3:54

1 Answer 1

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Pay close attention to the following fragment of the proof: "These inequalities hold for all $\lambda \gt 1$." In particular, consider $\lambda$ arbitrarily close to the value one.

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  • $\begingroup$ How do we know $\lambda$ arbitrarily close to the value 1 ? $\endgroup$ Oct 24, 2019 at 3:53
  • $\begingroup$ Plz note that i don't have knowledge to infer from "These inequalities hold for all $\lambda \gt 1$.", if possible elaborate. $\endgroup$ Oct 24, 2019 at 3:55
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    $\begingroup$ You don't need much "knowledge" for this. It's trivial. $\endgroup$
    – amsmath
    Oct 24, 2019 at 3:56
  • $\begingroup$ Go back to the second and third sentences of the proof. It's simply a matter of decoding what is assumed. You decode using the definitions. For example, you use the definition of a definite integral that is improper because it has infinity as the upper bound. $\endgroup$ Oct 24, 2019 at 3:59
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    $\begingroup$ @amsmath: using the label "trivial" doesn't automatically make it trivial for the student. You're describing your own evaluation of it. Also, remember that people stumble over molehills, not mountains. So, if somebody is stumbling and it is a trivial molehill, then it's a normal situation (as opposed to orthonormal or paranormal maybe). $\endgroup$ Oct 24, 2019 at 4:06

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