Tính \(\mathop {\lim \to {1^ - }} + 3x - {x - 1}

* Hướng dẫn giải

Khi \(x \to {1^ - }\) thì \(x < 1 \Rightarrow x - 1 < 0\) nên \(\left| {x - 1} \right| =  - \left( {x - 1} \right)\).

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} + 3x - 4}}{{\left| {x - 1} \right|}}\\ = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x - 1} \right)\left( {x + 4} \right)}}{{ - \left( {x - 1} \right)}}\\ = \mathop {\lim }\limits_{x \to {1^ - }} \left( { - x - 4} \right) =  - 1 - 4 =  - 5.\end{array}\)