# Finding limit of function

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.

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

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?