Câu hỏi 13

 Ta có: \(\underset{x\to -2}{\mathop{\lim }}\,\frac{x+1}{{{\left( x+2 \right)}^{2}}}=\underset{x\to -2}{\mathop{\lim }}\,\frac{-2+1}{{{\left( -2+2 \right)}^{2}}}=-\infty .\)

Chọn A.