Cho lim x → 3 f ( x ) = − 2 . Khi đó

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Cho \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = - 2\). Khi đó

Giới hạn \(\mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{f\left( x \right)}} = \infty \).

ĐúngSai

Giới hạn \(\mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right)}}{{x - 3}}\) tồn tại hữu hạn.

ĐúngSai

\[\mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right)}}{{{{\left( {x - 3} \right)}^2}}} = - \infty \].

ĐúngSai

Giới hạn \(\mathop {\lim }\limits_{x \to 3} \frac{{{{\left( {x - 3} \right)}^2}}}{{f\left( x \right)}} = 0\).

ĐúngSai

Giải thích

a) \(\mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{f\left( x \right)}} = 0\) vì \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = - 2\) và \(\mathop {\lim }\limits_{x \to 3} \left( {x - 3} \right) = 0\).

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right)}}{{x - 3}} = \infty \) vì \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = - 2\) và \(\mathop {\lim }\limits_{x \to 3} \left( {x - 3} \right) = 0\).

c) \[\mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right)}}{{{{\left( {x - 3} \right)}^2}}} = - \infty \] vì \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = - 2\); \(\mathop {\lim }\limits_{x \to 3} {\left( {x - 3} \right)^2} = 0\) và \({\left( {x - 3} \right)^2} > 0\) khi \(x \to 3\).

d) \(\mathop {\lim }\limits_{x \to 3} \frac{{{{\left( {x - 3} \right)}^2}}}{{f\left( x \right)}} = 0\) vì \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = - 2\); \(\mathop {\lim }\limits_{x \to 3} {\left( {x - 3} \right)^2} = 0\).

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