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Find the following limit : $$ \lim_{t \to(\pi/2)^-} \log\left(\frac{2 t}{\pi}\right) \log(\cos(t))$$ The indeterminate form is $0 \times\infty$ $$ \lim_{t \to(\pi/2)^-} \frac{ \log(\cos(t))}{\frac{1}{\log(\frac{2 t}{\pi})}}$$ And now it is in form $\frac{\infty}{\infty}$, but l'Hospital's rule doesn't help me. Any help or hint would be appreciated.

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    $\begingroup$ Excuse me. What is $x \to\pi/2-0$? $\endgroup$
    – Sebastiano
    Dec 17 '19 at 21:53
  • $\begingroup$ $x \to \pi/2 - $ $\endgroup$
    – Mark
    Dec 17 '19 at 21:54
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    $\begingroup$ What does $x$ have to do with $t$? $\endgroup$ Dec 17 '19 at 21:55
  • $\begingroup$ IMHO is better $t \to(\pi/2)^-$ or $t \to \left(\frac{\pi}2\right)^{-}$. $\endgroup$
    – Sebastiano
    Dec 17 '19 at 21:55
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Hint:

Set $u=\frac\pi2-t$ and rewrite the function: $$\log\Bigl(\frac{2 t}{\pi}\Bigr)\log(\cos t)=\log\Bigl(1-\frac{2 u}{\pi}\Bigr)\log(\sin u).$$ Now, near $0$, we have $$\log\Bigl(1-\frac{2 u}{\pi}\Bigr)\sim-\frac{2 u}{\pi}, \qquad \sin u\sim u\quad\text{hence }\;\log(\sin u)\sim\log u,$$ so that $$\log\Bigl(1-\frac{2 u}{\pi}\Bigr)\log(\sin u)\sim_0 -\frac 2\pi u\log u.$$ Can you conclude now?

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  • $\begingroup$ For my humble opinion the answer is OK! :-) $\endgroup$
    – Sebastiano
    Dec 17 '19 at 22:00

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