Question

Salut, trebuie sa rezolv o limita fara sa folosesc regula lui l'Hôpital

lim >x-0 (2^cosx - 2)/(2^x-1)

Stiu ca raspunsul trebuie sa imi dea 0 dar nu stiu pasii.

Multumesc!

Answer 1

$\textbf{Limite remarcabile:}\\\\\lim\limits_{x\to 0}\dfrac{a^x-1}{x} = \ln a,\quad a>0\\ \\ \lim\limits_{x\to 0}\dfrac{\sin x}{x} = 1\\\\\\\\ \textbf{Rezolvare:}\\\\\lim\limits_{x\to 0}\dfrac{2^{\cos x}-2}{2^x-1}=\lim\limits_{x\to 0}\dfrac{2\cdot (2^{\cos x-1}-1)}{2^x-1} =\\ \\ =2\cdot \lim\limits_{x\to 0}\dfrac{2^{\cos x-1}-1}{\cos x-1}\cdot \lim\limits_{x\to 0}\dfrac{\cos x - 1}{2^x-1} = \\ \\ = 2\cdot\ln 2\cdot\lim\limits_{x\to 0}\dfrac{\cos x - 1}{2^x-1} =$

$= 2\ln 2\cdot\lim\limits_{x\to 0}\dfrac{x}{2^x-1}\cdot \lim\limits_{x\to 0}\dfrac{\cos x-1}{x} =\\ \\=2\ln 2\cdot \dfrac{1}{\ln 2}\cdot\lim\limits_{x\to 0}\dfrac{(\cos x-1)(\cos x +1)}{x(\cos x+1)} =\\ \\=2\cdot\lim\limits_{x\to 0}\dfrac{\cos^2 x-1}{x(\cos x+1)} = \\ \\ = 2\cdot \lim\limits_{x\to 0}\dfrac{\sin^2 x}{x(\cos x+1)} =\\ \\ = 2\cdot \lim\limits_{x\to 0}\dfrac{\sin x}{x} \cdot \lim\limits_{x\to 0} \dfrac{\sin x}{\cos x +1} = \\ \\= 2\cdot 1\cdot \dfrac{\sin 0}{\cos 0+1} =$

$= 2\cdot 1\cdot 0 = \\ \\ =\boxed{0}$