Cho hàm số f ( x ) = { 3 √ 4 x 2 + 8 − √ 8 x 2 + 4 x k h i x ≠ 0 0 k h i x = 0 . Giá trị của f′(0) bằng:
Ta có:
\[f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{4{x^2} + 8}} - \sqrt {8{x^2} + 4} }}{{{x^2}}}\]
\[ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{4{x^2} + 8}} - 2}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[{}]{{8{x^2} + 4}} - 2}}{{{x^2}}}\]
\[ = \mathop {\lim }\limits_{x \to 0} \frac{{4{x^2}}}{{\left( {{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4} \right)}} - \mathop {\lim }\limits_{x \to 0} \frac{{8{x^2}}}{{{x^2}\left( {{{\sqrt[{}]{{4{x^2} + 8}}}^2} + 2} \right)}}\]
\[ = \mathop {\lim }\limits_{x \to 0} \frac{4}{{{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4}} - \mathop {\lim }\limits_{x \to 0} \frac{8}{{\sqrt[{}]{{8{x^2} + 4}} + 2}} = \frac{1}{3} - 2 = \frac{{ - 5}}{3}\]