Câu hỏi:
\(\mathop {\lim }\limits_{x \to {2^2}} \dfrac{{2x - 5}}{{x - 2}}\) bằng
Phương pháp giải:
\(\mathop {\lim }\limits_{x \to {x_0}^ + } f\left( x \right) = L < 0;\mathop {\lim }\limits_{x \to {x_0}^ + } g\left( x \right) = 0\) và \(g\left( x \right) > 0\) khi \(x \to {x_0}^ + \) thì \(\mathop {\lim }\limits_{x \to x_0^ + } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = - \infty \)
Lời giải chi tiết:
\(\begin{array}{l}\mathop {\lim }\limits_{x \to {2^2}} 2x - 5 = - 1 < 0\\\mathop {\lim }\limits_{x \to {2^2}} \left( {x - 2} \right) = 0;\\x - 2 > 0\forall x > 2\end{array}\)
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^2}} \dfrac{{2x - 5}}{{x - 2}} = - \infty \)