Question

limita când x tinde la 0 din ln(2-3^x)totul supra sin 5x.cum pot aduce numitorul la o forma mai simpla pt a fi limita fundamentala?

Answer 1

[tex]\displaystyle \\ \lim_{x \to 0} \frac{ln(2-3^x)}{\sin5x} \\ \\ \text{Aplicam regula lui l'Hopital} \\ \\ \lim_{x \to 0} \frac{\ln(2-3^x)}{\sin5x} = \lim_{x \to 0} \frac{\ln'(2-3^x)}{\sin'5x}= \\ \\ =\lim_{x \to 0} \frac{ \frac{(2-3^x)'}{2-3^x} }{(-5x)'\cos x}= \lim_{x \to 0} \frac{ \frac{-3^x \ln3}{2-3^x} }{-5 \cos x}=\lim_{x \to 0} \frac{ \frac{3^x \ln3}{2-3^x} }{5 \cos x}= [/tex]


[tex]\displaystyle \\ =\lim_{x \to 0} \frac{3^x \ln3}{(2-3^x)\cdot 5 \cos x}=\\\\ =\frac{3^0 \ln3}{(2-3^0)\cdot 5 \cos 0}= \frac{1\cdot \ln3}{(2-1)\cdot 5 \cdot 1}= \boxed{\frac{\ln 3}{5}}[/tex]