Question

[Trần Hà Linh]

Tính các giới hạn 

a) \(\lim\limits_{x\rightarrow2}\dfrac{x+3}{x^2+x+4}=\dfrac{1}{2}\)

b) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2+5x+6}{x^2+3x}=\dfrac{1}{3}\)

Answer 1

[Hoàng Tử Hà]

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)