Giải giúp mình với Lim x²-√2x+3 x-1 ————— x+1 11/10/2021 Bởi Skylar Giải giúp mình với Lim x²-√2x+3 x-1 ————— x+1
Đáp án: \[\mathop {\lim }\limits_{x \to – 1} \dfrac{{{x^2} – \sqrt {2x + 3} }}{{x + 1}} = – 3\] Giải thích các bước giải: Ta có: \(\begin{array}{l}\mathop {\lim }\limits_{x \to – 1} \dfrac{{{x^2} – \sqrt {2x + 3} }}{{x + 1}}\\ = \mathop {\lim }\limits_{x \to – 1} \dfrac{{\left( {{x^2} – 1} \right) + \left( {1 – \sqrt {2x + 3} } \right)}}{{x + 1}}\\ = \mathop {\lim }\limits_{x \to – 1} \left[ {\dfrac{{{x^2} – 1}}{{x + 1}} + \dfrac{{1 – \sqrt {2x + 3} }}{{x + 1}}} \right]\\ = \mathop {\lim }\limits_{x \to – 1} \left[ {\dfrac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{x + 1}} + \dfrac{{\left( {1 – \sqrt {2x + 3} } \right)\left( {1 + \sqrt {2x + 3} } \right)}}{{\left( {x + 1} \right)\left( {1 + \sqrt {2x + 3} } \right)}}} \right]\\ = \mathop {\lim }\limits_{x \to – 1} \left[ {\left( {x – 1} \right) + \dfrac{{{1^2} – \left( {2x + 3} \right)}}{{\left( {x + 1} \right)\left( {1 + \sqrt {2x + 3} } \right)}}} \right]\\ = \mathop {\lim }\limits_{x \to – 1} \left[ {\left( {x – 1} \right) + \dfrac{{ – 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {1 + \sqrt {2x + 3} } \right)}}} \right]\\ = \mathop {\lim }\limits_{x \to – 1} \left[ {\left( {x – 1} \right) – \dfrac{2}{{1 + \sqrt {2x + 3} }}} \right]\\ = \left( { – 1 – 1} \right) – \dfrac{2}{{1 + \sqrt {2.\left( { – 1} \right) + 3} }}\\ = – 3\end{array}\) Bình luận
Đáp án:
\[\mathop {\lim }\limits_{x \to – 1} \dfrac{{{x^2} – \sqrt {2x + 3} }}{{x + 1}} = – 3\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to – 1} \dfrac{{{x^2} – \sqrt {2x + 3} }}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to – 1} \dfrac{{\left( {{x^2} – 1} \right) + \left( {1 – \sqrt {2x + 3} } \right)}}{{x + 1}}\\
= \mathop {\lim }\limits_{x \to – 1} \left[ {\dfrac{{{x^2} – 1}}{{x + 1}} + \dfrac{{1 – \sqrt {2x + 3} }}{{x + 1}}} \right]\\
= \mathop {\lim }\limits_{x \to – 1} \left[ {\dfrac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{x + 1}} + \dfrac{{\left( {1 – \sqrt {2x + 3} } \right)\left( {1 + \sqrt {2x + 3} } \right)}}{{\left( {x + 1} \right)\left( {1 + \sqrt {2x + 3} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to – 1} \left[ {\left( {x – 1} \right) + \dfrac{{{1^2} – \left( {2x + 3} \right)}}{{\left( {x + 1} \right)\left( {1 + \sqrt {2x + 3} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to – 1} \left[ {\left( {x – 1} \right) + \dfrac{{ – 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {1 + \sqrt {2x + 3} } \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to – 1} \left[ {\left( {x – 1} \right) – \dfrac{2}{{1 + \sqrt {2x + 3} }}} \right]\\
= \left( { – 1 – 1} \right) – \dfrac{2}{{1 + \sqrt {2.\left( { – 1} \right) + 3} }}\\
= – 3
\end{array}\)