Cho hàm số f(x) có lim khi x → 3 của f(x) = −2 và g(x) = (x − 3)^3. Xét tính đúng sai của các khẳng định sau:

Hướng dẫn giải:

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

a) Đúng. \(\mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ - }} {\left( {x - 3} \right)^3} = 0\)

b) Sai. Ta có:\(\mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right) = 0,\forall x > 3\).

Do đó \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{g\left( x \right)}} = + \infty \).

c) Đúng. \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to {3^ + }} \left( {f\left( x \right) \cdot \frac{1}{{g\left( x \right)}}} \right)\)

Do \(\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} f\left( x \right) = - 2;\,\,\,\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{g\left( x \right)}} = + \infty \)

nên \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to {3^ + }} \left( {f\left( x \right) \cdot \frac{1}{{g\left( x \right)}}} \right) = - \infty \)

d) Sai. Ta có: \[\mathop {\lim }\limits_{x \to {3^ + }} \left[ {\frac{1}{{f\left( x \right) \cdot g\left( x \right)}}} \right] = \mathop {\lim }\limits_{x \to {3^ + }} \left[ {\frac{1}{{f\left( x \right)}} \cdot \frac{1}{{g\left( x \right)}}} \right]\]

Do \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{f\left( x \right)}} = - \frac{1}{2};\,\,\mathop {\lim }\limits_{x \to {3^ + }} \frac{1}{{g\left( x \right)}} = + \infty \) nên \(\mathop {\lim }\limits_{x \to {3^ + }} \left[ {\frac{1}{{f\left( x \right) \cdot g\left( x \right)}}} \right]\)

\( = \mathop {\lim }\limits_{x \to {3^ + }} \left[ {\frac{1}{{f\left( x \right)}} \cdot \frac{1}{{g\left( x \right)}}} \right] = - \infty \).