Question

Calculati
$\lim_{x \to 0, \ x \ \textgreater \ 0} \frac{arctg\sqrt[6]{x}}{arctg\sqrt[5]{x}}$

Va rog sa fie cat de cat detaliat.. mie nu imi iese (rezultatul este ∞ )

Answer 1

Răspuns:

Explicație pas cu pas:

Rezolvarea in imaginea de mai jos.

Answer 2

$l = \lim\limits_{x\searrow 0}\dfrac{\arctan\sqrt[6]{x}}{\arctan\sqrt[5]x}\\ \\ \\\sqrt[5]{x} = x^{\frac{1}{5}} = x^{\frac{6}{30}} \\ \\ \sqrt[6]x = x^{\frac{1}{6}} = x^{\frac{5}{30}}\\ \\\\ \text{Fac schimbarea de variabila: }\,\,x^{\frac{1}{30}} = t\\\\ x\searrow 0 \Rightarrow x^{\frac{1}{30}} \searrow 0 \Rightarrow t\searrow 0\\ \\ \\ l=\lim\limits_{t\searrow 0}\dfrac{\arctan (t^5)}{\arctan (t^6)} = \lim\limits_{t\searrow 0}\Big(\dfrac{\arctan (t^5)}{t^5}\cdot \dfrac{t^6}{\arctan (t^6)}\cdot \dfrac{1}{t}\Big) =$

$= \lim\limits_{t\searrow 0}\Big(\dfrac{\arctan (t^5)}{t^5}\Big)\cdot \lim\limits_{t\searrow 0}\Big(\dfrac{t^6}{\arctan(t^6)}\Big)\cdot \lim\limits_{t\searrow 0}\dfrac{1}{t} = \\\\ = 1\cdot 1\cdot \lim\limits_{t\searrow 0}\dfrac{1}{t} = \dfrac{1}{0_+} = \boxed{+\infty}$

$\\\\\text{Limit\u{a} remarcabil\u{a}:} \\ \\ \lim\limits_{x\to x_0}\dfrac{\arctan \Big(u(x)\Big)}{u(x)} = 1,\quad \mathrm{c\hat and}\,\,u(x)\to 0$