Tính giới hạn sau: lim x tiến đến 1 ((căn bậc hai của (2x^2+7)−3(căn bậc ba của (3x+5))

* Hướng dẫn giải

$\underset{\mathrm{x}\to 1}{\lim }\left(\frac{\sqrt{2{\mathrm{x}}^2+7}-3\sqrt[3]{3\mathrm{x}+5}+\sqrt[4]{3\mathrm{x}+78}}{{\mathrm{x}}^2-1}\right)$

$=\underset{\mathrm{x}\to 1}{\lim }(\frac{\sqrt{2{\mathrm{x}}^2+7}-3}{{\mathrm{x}}^2-1}-3\frac{\sqrt[3]{3\mathrm{x}+5}-2}{{\mathrm{x}}^2-1}$$+\frac{\sqrt[4]{3\mathrm{x}+78}-3}{{\mathrm{x}}^2-1})$

$=\underset{\mathrm{x}\to 1}{\lim }\frac{1}{{\mathrm{x}}^2-1}(\frac{2{\mathrm{x}}^2-2}{\sqrt{2{\mathrm{x}}^2+7}+3}$$-3\frac{3\mathrm{x}-3}{\sqrt[3]{{\left(3\mathrm{x}+5\right)}^2}+2\sqrt[3]{3\mathrm{x}+5}+4}$$+\frac{3\mathrm{x}-3}{\left(\sqrt[4]{3\mathrm{x}+78}+3\right)\left(\sqrt{3\mathrm{x}+78}+9\right)})$

$=\underset{\mathrm{x}\to 1}{\lim }(\frac{2}{\sqrt{2{\mathrm{x}}^2+7}+3}$$-\frac{9}{\left(\sqrt[3]{{\left(3\mathrm{x}+5\right)}^2}+2\sqrt[3]{3\mathrm{x}+5}+4\right)\left(\mathrm{x}+1\right)}$$+\frac{3}{\left(\mathrm{x}+1\right)\left(\sqrt[4]{3\mathrm{x}+78}+3\right)\left(\sqrt{3\mathrm{x}+78}+9\right)})$

$=\frac{-1}{36}.$