Question

Urgent va rog....................

Answer 1

Răspuns:

Explicație pas cu pas:

$a) -1\leq \sin\dfrac{1}{x-1}\leq 1\\-\ln x\leq \sin\dfrac{1}{x-1}\cdot\ln x\leq \ln x\\\text{Deoarece }\displaystyle\lim_{x\to1} (-\ln x)=\lim_{x\to1}\ln x=0,\texttt{ din criteriul clestelui}\\\texttt{rezulta ca }\lim_{x\to 1}\sin\dfrac{1}{x-1}\cdot \ln x=0$

$b)-1\leq\cos\dfrac{1}{x-\pi}\leq 1\\-\sin x\leq\sin x\cdot \cos\dfrac{1}{x-\pi}\leq\sin x\\\texttt{Se aplica acelasi principiu ca la subpunctul anterior, deci }\\\displaystyle\lim_{x\to\pi}\sin x\cdot\cos\dfrac{1}{x-\pi}=0\\\\\\c)\lim_{x\to\infty}\dfrac{2^x-1}{2^x+1}=\lim_{x\to\infty}\dfrac{2^x\left(1-\frac{1}{2^x}\right)}{2^x\left(1+\frac{1}{2^x}\right)}=\dfrac{1-0}{1+0}=1\\d)\lim_{x\to\infty}\left(x+\arcsin\dfrac{1}{x}\right)=\infty+\arcsin 0=\infty$