Tính các giới hạn sau:         \(a)\mathop {\lim }\limits_{x \to  + \infty } \frac{{x + 1}}{{2x + 1}}{\rm{  &

* Hướng dẫn giải

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{x + 1}}{{2x + 1}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{1 + \frac{1}{x}}}{{2 + \frac{1}{x}}} = \frac{1}{2}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ + }} \left( {3x - 1} \right) = 5 > 0;\mathop {\lim }\limits_{x \to {2^ + }} \left( {x - 2} \right) = 0\\
x \to {2^ + } \Rightarrow x - 2 > 0\\
 \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} \frac{{3x - 1}}{{x - 2}} =  + \infty 
\end{array}\)