Đáp án:
\[\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {{x^2} - 5} + 2{x^2} - 20}}{{x - 3}} = \frac{{27}}{2}\]
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {{x^2} - 5} + 2{x^2} - 20}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {\sqrt {{x^2} - 5} - 2} \right) + \left( {2{x^2} - 18} \right)}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{{\frac{{{x^2} - 5 - {2^2}}}{{\sqrt {{x^2} - 5} + 2}} + 2.\left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{{\frac{{{x^2} - 9}}{{\sqrt {{x^2} - 5} + 2}} + 2.\left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \dfrac{{\frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\sqrt {{x^2} - 5} + 2}} + 2.\left( {x - 3} \right)\left( {x + 3} \right)}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to 3} \left[ {\frac{{x + 3}}{{\sqrt {{x^2} - 5} + 2}} + 2.\left( {x + 3} \right)} \right]\\
= \frac{{3 + 3}}{{\sqrt {{3^2} - 5} + 2}} + 2.\left( {3 + 3} \right)\\
= \frac{6}{4} + 12\\
= \frac{{27}}{2}
\end{array}\)
