Giải thích các bước giải:
\(\begin{array}{l}
f,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {6x + 3} + 2{x^2} - 5x}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {6x + 3} - \left( {x + 2} \right)} \right) + \left( {2{x^2} - 4x + 2} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{6x + 3 - {{\left( {x + 2} \right)}^2}}}{{\sqrt {6x + 3} + x + 2}} + 2{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{ - {{\left( {x - 1} \right)}^2}}}{{\sqrt {6x + 3} + x + 2}} + 2{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{ - 1}}{{\sqrt {6x + 3} + x + 2}} + 2} \right]\\
= \frac{{ - 1}}{{\sqrt {6.1 + 3} + 1 + 2}} + 2\\
= \frac{{ - 1}}{6} + 2 = \frac{{11}}{6}\\
g,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[4]{{4x - 3}} - 1}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\sqrt {4x - 3} - 1}}{{\sqrt[4]{{4x - 3}} + 1}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {\sqrt {4x - 3} - 1} \right)\left( {\sqrt {4x - 3} + 1} \right)}}{{\left( {\sqrt[4]{{4x - 3}} + 1} \right)\left( {\sqrt {4x - 3} + 1} \right)}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{4x - 4}}{{\left( {\sqrt[4]{{4x - 3}} + 1} \right)\left( {\sqrt {4x - 3} + 1} \right)}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{4}{{\left( {\sqrt[4]{{4x - 3}} + 1} \right)\left( {\sqrt {4x - 3} + 1} \right)}}\\
= \frac{4}{{\left( {\sqrt[4]{{4.1 - 3}} + 1} \right)\left( {\sqrt {4.1 - 3} + 1} \right)}}\\
= \frac{4}{{2.2}}\\
= 1
\end{array}\)
