Cho 2 hàm số f(x) = x/|x − 2| và g(x) = (x − 1)/(x^2). Xét tính đúng sai của các khẳng định sau:

Hướng dẫn giải:

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

a) Đúng. Vì x $\to$ 2⁺ nên x – 2 > 0 Û |x – 2| = x – 2.

Do đó, \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \frac{x}{{x - 2}}\).

b) Sai. Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) = - 1 < 0\\\mathop {\lim }\limits_{x \to 0} {x^2} = 0\\{x^2} > 0\,\,khi\,\,x \to 0\end{array} \right.\, \Rightarrow \mathop {\lim }\limits_{x \to 0} g\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x - 1}}{{{x^2}}} = - \infty \).

c) Đúng. Khi x $\to$ 2^- thì |x – 2| =-x + 2

Vì \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {2^ - }} x = 2 > 0\\\mathop {\lim }\limits_{x \to {2^ - }} \left( { - x + 2} \right) = 0\\ - x + 2 > 0\,\,khi\,\,x \to {2^ - }\end{array} \right.\)nên\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{ - x + 2}} = + \infty \)

d) Đúng. Ta có: \(\mathop {\lim }\limits_{x \to {0^ + }} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{x}{{\left| {x - 2} \right|}} \cdot \frac{{x - 1}}{{{x^2}}}} \right] = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{1}{x} \cdot \frac{{x - 1}}{{\left| {x - 2} \right|}}} \right]\)

Lại có: \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{x - 1}}{{\left| {x - 2} \right|}} = - \frac{1}{2}\) và \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} = + \infty \) nên \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} \cdot \frac{{x - 1}}{{\left| {x - 2} \right|}} = - \infty \).