Question

Tính \(\mathop {\lim }\limits_{x \to  - {4^ + }} \left( {\sqrt {x + 4}  + x} \right)\)

Answer 1

Với dãy số \(\left( {{x_n}} \right)\) bất kì \({x_n} >  - 4\) và \({x_n} \to  - 4,\) ta có:

\(\begin{array}{c}\mathop {\lim }\limits_{{x_n} \to  - {4^ + }} \left( {\sqrt {{x_n} + 4}  + {x_n}} \right) = \mathop {\lim }\limits_{{x_n} \to  - {4^ + }} \sqrt {{x_n} + 4}  + \mathop {\lim }\limits_{{x_n} \to  - {4^ + }} {x_n} = \sqrt {\mathop {\lim }\limits_{{x_n} \to  - {4^ + }} \left( {{x_n} + 4} \right)}  + \left( { - 4} \right)\\ = \sqrt {\mathop {\lim }\limits_{{x_n} \to  - {4^ + }} {x_n} + 4}  - 4 = \sqrt { - 4 + 4}  - 4 =  - 4\end{array}\)

Vậy \(\mathop {\lim }\limits_{x \to  - {4^ + }} \left( {\sqrt {x + 4}  + x} \right) =  - 4\)