Question

Am nevoie de rezolvare

Answer 1

[tex] \displaystyle \lim_{x \to 0} \frac{3^x-4^x}{2^x-4^x}= \lim_{x \to 0} \frac{(3^x-4^x)'}{(2^x-4^x)'}=\\ = \lim_{x \to 0} \frac{ln3\cdot 3^x-ln4\cdot 4^x}{ln2 \cdot 2^x-ln4\cdot 4^x}=\\ =\frac{ln3 -ln4 }{ln2 -ln4 }=\\ =\frac{ \frac{log_23}{log_2e} - \frac{log_24}{log_2e} }{ \frac{log_22}{log_2e} - \frac{log_24}{log_2e} }=\\ = \frac{log_23-2}{1-2} =2-log_23[/tex]Scrie răspunsul tău aici