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afasiask7xg

Answered question

2022-04-01

I am trying to find this limit but I failed.

lim_{x \to \infty} (2 \arctan x - π) \ln x

Answer & Explanation

Marcos Boyer

Beginner2022-04-02Added 12 answers

(2 \arctan (x) - π) \ln (x) = - 2 (\frac{π}{2} - \arctan (x)) \ln (x) = - 2 \arctan (\frac{1}{x}) \ln (x)

lim_{x \to \infty} (2 \arctan (x) - π) \ln (x) = - 2 lim_{x \to \infty} \arctan (\frac{1}{x}) \ln (x) = 2 lim_{x \to 0^{+}} \arctan (x) \ln (x)

lim_{x \to 0^{+}} \arctan (x) \ln (x) = lim_{x \to 0^{+}} \frac{\arctan (x)}{\frac{1}{\ln (x)}} = lim_{x \to 0^{+}} \frac{\frac{1}{1 + x^{2}}}{- \frac{1}{\ln^{2} (x)} \frac{1}{x}} = - lim_{x \to 0^{+}} \frac{x \ln^{2} (x)}{1 + x^{2}} = 0

Hence,

lim_{x \to \infty} (2 \arctan (x) - π) \ln (x) = 0

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