Cho Lim f(x) - 10/x - 1 = 5. Giới hạn

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Cho \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{x - 1}} = 5\). Giới hạn \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt {4f\left( x \right) + 9} + 3} \right)}}\) bằng

\(1\).

\(10\).

\(2\).

\(\frac{5}{3}\).

Giải thích

Chọn A

\(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt {4f\left( x \right) + 9} + 3} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\sqrt x - 1}}.\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {4f\left( x \right) + 9} + 3}}\)

Xét \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\sqrt x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left[ {f\left( x \right) - 10} \right]\left( {\sqrt x + 1} \right)}}{{x - 1}} = 5\left( {1 + 1} \right) = 10\)

Xét \[\mathop {\lim }\limits_{x \to 1} \sqrt {4f\left( x \right) + 9} = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {4\frac{{\left( {f\left( x \right) - 10 + 10} \right)}}{{x - 1}}\left( {x - 1} \right) + 9} } \right)\]

\[ = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {4\left[ {\frac{{f\left( x \right) - 10}}{{x - 1}} + \frac{{10}}{{x - 1}}} \right]\left( {x - 1} \right) + 9} } \right)\]

\[ = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {4\left( {5 + \frac{{10}}{{x - 1}}} \right)\left( {x - 1} \right) + 9} } \right)\]

\[ = \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {20\left( {x - 1} \right) + 40 + 9} } \right) = \sqrt {49} = 7\]

Suy ra \[\mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {4f\left( x \right) + 9} + 3}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{7 + 3}} = \frac{1}{{10}}\]

\(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 10}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt {4f\left( x \right) + 9} + 3} \right)}} = 10.\frac{1}{{10}} = 1\)