[tex]\displaystyle a)\,\, \lim\limits_{x\to \infty} \frac{2\sqrt{x+1}}{\sqrt{3+x}} = \lim\limits_{x\to \infty} \frac{2\sqrt{x\left(1+\frac{1}{x}\right)}}{\sqrt{x\left(3+\frac{1}{x}\right)} }= \lim\limits_{x\to \infty} \frac{2\sqrt{x\left(1+\frac{1}{x}\right)}}{\sqrt{x\left(3+\frac{1}{x}\right)}} = [/tex]
[tex]\displaystyle = \lim\limits_{x\to \infty} \frac{2\sqrt{x} \sqrt{1+\frac{1}{x}}}{\sqrt{x}\sqrt{\frac{3}{x}+1}} = \lim\limits_{x\to \infty} \frac{2\sqrt{1+\frac{1}{x}}}{\sqrt{\frac{3}{x}+1}} = \frac{2\sqrt{1+\frac{1}{\infty}}}{\sqrt{\frac{3}{\infty}+1}} = \frac{2\sqrt{1+0}}{\sqrt{0+1}} = \frac{2\sqrt{1}}{\sqrt{1}} =2[/tex]
Restul se rezolvă tot prin scoaterea factorului forțat.
Observăm că limitele de tipul acesta sunt egale cu fracția formată din x-șii care au puterea cea mai mare împreună cu coeficienții lor.
[tex]\displaystyle b)\,\, \lim\limits_{x\to -\infty} \frac{\sqrt{x^2+x}}{\sqrt{4x^2+3}} = ...= \lim\limits_{x\to -\infty} \frac{\sqrt{x^2}}{\sqrt{4x^2}} = \frac{1}{2}[/tex]
[tex]\displaystyle c)\,\, \lim\limits_{x\to \infty} \frac{x+\sqrt{x}}{3x+2\sqrt{x}+1} = ...= \lim\limits_{x\to \infty} \frac{x}{3x} = \frac{1}{3}[/tex]
[tex]\displaystyle d)\,\, \lim\limits_{x\to \infty} \frac{\sqrt[3]{x}+x}{2x+3} = ...= \lim\limits_{x\to \infty} \frac{x}{2x} = \frac{1}{2}[/tex]
[tex]\displaystyle e)\,\, \lim\limits_{x\to \infty} \frac{\sqrt{x^2+1}+2x}{2\sqrt{x^2-1}+x} = ...= \lim\limits_{x\to \infty} \frac{x+2x}{2x+x} = \lim\limits_{x\to \infty} \frac{3x}{3x} = 1[/tex]
[tex]\displaystyle f)\,\, \lim\limits_{x\to \infty} \frac{\sqrt{x}+2\sqrt{x+1}}{3\sqrt{x-1}+\sqrt{4x+1}} = ...= \lim\limits_{x\to \infty} \frac{\sqrt{x}+2\sqrt{x}}{3\sqrt{x}+\sqrt{4}\sqrt{x}} = \lim\limits_{x\to \infty} \frac{3\sqrt{x}}{5\sqrt{x}} = \frac{3}{5}[/tex]
[tex]\displaystyle g)\,\, \lim\limits_{x\to -\infty} \frac{3x-1}{\sqrt{9x^2-x+7}} = ...= \lim\limits_{x\to -\infty} \frac{3x}{\sqrt{9}\sqrt{x^2}} = \lim\limits_{x\to -\infty} \frac{3x}{3x} = 1[/tex]
[tex]\displaystyle h)\,\, \lim\limits_{x\to -\infty} \frac{2x^2-3x+5}{3x-4} = ...= \lim\limits_{x\to -\infty} \frac{2x^2}{3x} = \lim\limits_{x\to -\infty} \frac{2x}{3} =\frac{2\cdot (-\infty)}{3} = -\infty[/tex]
