Kết quả của giới hạn $\mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{\sqrt {x + 2} }}{{\sqrt {x - 2} }}$ là:

$- \infty$

$\dfrac{5}{2}$.

$+ \infty$

$2$

7

3

$- 7$

$- 3$

3

$- \infty$

2

$+ \infty .$

1

$- \infty$.

0

$+ \infty$.

$\mathop {\lim }\limits_{x \to {x_0}} x = {x_0}$  

$\mathop {\lim }\limits_{x \to {x_0}} x = 1$

$\mathop {\lim }\limits_{x \to {x_0}} c = {x_0}$

$\mathop {\lim }\limits_{x \to {x_0}} x = 0$

$\dfrac{1}{4}.$

$\dfrac{1}{2}.$

$\dfrac{1}{3}.$

$\dfrac{1}{5}.$

$\mathop {\lim }\limits_{x \to {x_0}} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{L}{M}$

$\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right).g\left( x \right)} \right] = L.M$

$\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) - g\left( x \right)} \right] = L - M$    

$\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = L + M$