CALCOLO DI LIMITI CON I LIMITI NOTEVOLI

E/O CONFRONTO DI INFINITI O INFINITESIMI

$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{\sqrt{x}+\sqrt[3]{x}}{2+\sqrt{x}}$$
$$\underset{x\to 0^{\text{+}}}{\lim }\frac{\sqrt{1-\cos x}}{x}$$
$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{2x+1}{2x+3}\right)}^{x-1}$$
$$\underset{x\to \sqrt{2}}{\lim }\frac{x-\sqrt{2}}{x^2-2}$$
$$\underset{x\to 0}{\lim }\frac{{\log }_2(1+x)}{3x}$$
$$\underset{x\to 1}{\lim }\frac{2x^2-5x+3}{3x^3+7x^2-4x-6}$$
$$\underset{x\to +\mathrm{\infty }}{\lim }{\left(\frac{1}{x}\right)}^{-\frac{1}{1+\ln x}}$$
$$\underset{x\to +\mathrm{\infty }}{\lim }(\sqrt{x+1}-\sqrt{x^2-4})$$
$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{\pi x+\pi }{\pi x-\pi }\right)}^{\frac{\pi }{2}x+\frac{\pi }{3}}$$
$$\underset{x\to 0}{\lim }\frac{\sin 3x+\sin x}{x}$$
$$\underset{x\to -3}{\lim }\frac{x^3-7x+6}{x^2+8x+15}$$
$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{3x+5}{3x+3}\right)}^{2x+1}$$
$$\underset{x\to 1}{\lim }\frac{x^4-1}{x^3-x^2}$$
$$\underset{x\to -\frac{\pi }{4}}{\lim }\frac{\sqrt{2}+2\sin x}{\pi +4x}$$
$$\underset{x\to 0}{\lim }\frac{\ln (3x^2+1)}{x^2}$$
$$\underset{x\to 1}{\lim }\frac{\sqrt{x+1}-\sqrt{2}}{\sqrt{x+8}-3}$$
$$\underset{x\to +\mathrm{\infty }}{\lim }{\left(\frac{1}{2x}\right)}^{\frac{1}{\ln 3x}}$$
$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{1+\tan x}{1-\tan x}$$
$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{3x+7}{3x+6}\right)}^{6x+7}$$
$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{e^{2x}-2e^x-1}{3+e^{2x+1}}$$