Tính các giới hạn sau:a) \(\mathop {\lim }\limits_{x \to 1} \frac{{2 - x - {x^2}}}{{x - 1}}\)b) \(\mathop {\lim }\limits_{x \to {3^ + }} \fr

* Hướng dẫn giải

a) \(\mathop {\lim }\limits_{x \to 1} \frac{{2 - x - {x^2}}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{( - x - 2)(x - 1)}}{{(x - 1)}} = \mathop {\lim }\limits_{x \to 1} ( - x - 2) =  - 3\)

b) \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x + 2}}{{x - 3}} =  + \infty \) vì \(\left\{ \begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ + }} \left( {x + 2} \right) = 5\\
\mathop {\lim }\limits_{x \to {3^ + }} \left( {x - 3} \right) = 0\\
x - 3 > 0\,\,\,khi\,\,x \to {3^ + }
\end{array} \right.\)