Cho I = lim x → 0 2 ( √ 3 x + 1 − 1 ) x và J = lim x → − 1 (x^ 2 − x − 2)/( x + 1) . Tính I − J .

Ta có

\(I = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {3x + 1}  - 1} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{6x}}{{x\left( {\sqrt {3x + 1}  + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{6}{{\sqrt {3x + 1}  + 1}} = 3\).

\(J = \mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} - x - 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {x - 2} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \left( {x - 2} \right) =  - 3\).

Khi đó \(I - J = 6\).