Răspuns:
Explicație pas cu pas:
Prima parte am rezolvat-o deja . In schimb, pentru a doua:
[tex]e)\displaystyle\lim_{x\to\infty}(x+1-\sqrt x)=\lim_{x\to\infty}x\left(1+\dfrac{1}{x}-\dfrac{\sqrt x}{x}\right)=\infty(1+0-0)=\infty\\f)-1\leq\sin x\leq 1|+x^2\\x^2-1\leq x^2+\sin x\leq x^2+1|:x\\\dfrac{x^2-1}{x}\leq\dfrac{x^2+\sin x}{x}\leq\dfrac{x^2+1}{x}\\\texttt{Din teorema clestelui rezulta ca }\lim_{x\to\infty}\dfrac{x^2+\sin x}{x}=\infty[/tex]
[tex]h)-1\leq\sin x\leq 1|+3\\2\leq 3+\sin x\leq 4|\cdot \ln x\\2\ln x\leq(3+\sin x)\ln x\leq 4\ln x\\\texttt{Din criteriul clestelui rezulta ca }\displaystyle \lim_{x\to\infty}(3+\sin x)\ln x=\infty\\i)-1\leq\sin x\leq 1|\cdot(-1)\\1\geq -\sin x\geq -1|+\dfrac{x^2}{x+1}\\\dfrac{x^2}{x+1}+1\geq \dfrac{x^2}{x+1}-\sin x\geq \dfrac{x^2}{x+1}-1\\\texttt{Din criteriul clestelui rezulta ca }\displaystyle\lim_{x\to\infty}\left(\dfrac{x^2}{x+1}-\sin x\right)=\infty[/tex]
