lim x → 2 [ 5 f ( x ) ] = − ∞ .

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PHẦN II. TRẮC NGHIỆM ĐÚNG – SAI

Cho hai hàm số y = f(x); y = g(x) thỏa mãn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5\) và \(\mathop {\lim }\limits_{x \to 2} g\left( x \right) = + \infty \).

a) \(\mathop {\lim }\limits_{x \to 2} \left[ {5f\left( x \right)} \right] = - \infty \).

b)\(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right).g\left( x \right)} \right] = + \infty \).

c)\[\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = + \infty \].

d) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) - 1} - 2}}{{f\left( x \right) - 5}} = \frac{1}{4}\).

Nhập câu trả lời ở đây

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Giải thích

a) \(\mathop {\lim }\limits_{x \to 2} \left[ {5f\left( x \right)} \right] = 5\mathop {\lim }\limits_{x \to 2} f\left( x \right) = 5.5 = 25\).

b) \(\mathop {\lim }\limits_{x \to 2} \left[ {f\left( x \right).g\left( x \right)} \right] = + \infty \).

c) \[\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right)}}{{g\left( x \right)}} = 0\].

d) \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {f\left( x \right) - 1} - 2}}{{f\left( x \right) - 5}} = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 5}}{{\left( {\sqrt {f\left( x \right) - 1} + 2} \right)\left( {f\left( x \right) - 5} \right)}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{1}{{\sqrt {f\left( x \right) - 1} + 2}} = \frac{1}{4}\).

Đáp án: a) Sai; b) Đúng; c) Sai; d) Đúng.