Question

Sa se calculeze limita:

$\lim_{x \to \i0} \frac{ln(1+x^{2018})-ln^{2018}(1+x) }{x^{2018} }$

Answer 1

$l=\lim\limits_{x\to 0}\dfrac{\ln(1+x^{2018})-\ln^{2018}(1+x)}{x^{2018}}\\ \\l = \lim\limits_{x\to 0}\dfrac{\ln(1+x^{2018})}{x^{2018}}-\lim\limits_{x\to 0}\dfrac{\ln^{2018}(1+x)}{x^{2018}} \\ \\l = \lim\limits_{x\to 0}\dfrac{\ln(1+x^{2018})}{x^{2018}}- \lim\limits_{x\to 0}\Bigg[\dfrac{\ln(1+x)}{x}\Bigg]^{2018}\\ \\ l= 1-1 \\ \\ \Rightarrow \boxed{l = 0}$

M-am folosit de limita remarcabilă:

$\lim\limits_{x \to x_0}\dfrac{\ln\Big(1+u(x)\Big)}{u(x)} = 1,\quad \text{cand }u(x) \to 0$