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Решите задачу:

$\large \displaystyle \lim_{x \to 0}\frac{\ln {\rm tg}(\frac{\pi}{4}+ax)}{\sin bx}=\lim_{x \to 0}\frac{\ln\frac{1+{\rm tg}ax}{1-{\rm tg}ax}}{\sin bx}=\lim_{x \to 0}\frac{\ln(1+{\rm tg}ax)-\ln(1-{\rm tg}ax)}{\sin bx}=\\ \\ \\ =\lim_{x \to 0}\frac{\ln(1+{\rm tg}ax)}{\sin bx}-\lim_{x \to 0}\frac{\ln(1-{\rm tg}ax)}{\sin bx}=\lim_{x \to 0}\frac{{\rm tg}ax}{\sin bx}+\lim_{x \to 0}\frac{{\rm tg}ax}{\sin bx}=\\ \\ \\ =2\lim_{x \to 0}\frac{{\rm tg}ax}{\sin bx}=2\lim_{x \to 0}\frac{ax}{bx}=\frac{2a}{b}$