Tìm được các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to {2^ + }} (\sqrt {x + 2} - 1) = 1\);

b) \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{4x - 3}}{{x - 1}} = + \infty \);

c) \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{1}{{x - 2}} - \frac{1}{{{x^2} - 4}}} \right) = - \infty \);

d) \(\mathop {\lim }\limits_{x \to - {1^ - }} \frac{{|x + 1|}}{{{x^2} - 1}} = - \infty \).

a) \(\mathop {\lim }\limits_{x \to {2^ + }} (\sqrt {x + 2} - 1) = \sqrt {2 + 2} - 1 = 1\).

b) \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{4x - 3}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {(4x - 3) \cdot \frac{1}{{x - 1}}} \right] = + \infty \) vì \(\mathop {\lim }\limits_{x \to {1^ + }} (4x - 3) = 1,\mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{x - 1}} = + \infty \).

c)

\({\rm{ }}\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to {2^ - }} \left( {\frac{1}{{x - 2}} - \frac{1}{{{x^2} - 4}}} \right)}&{ = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 2 - 1}}{{(x - 2)(x + 2)}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 1}}{{(x - 2)(x + 2)}}}\\{}&{ = \mathop {\lim }\limits_{x \to {2^ - }} \left[ {\frac{{x + 1}}{{x + 2}} \cdot \frac{1}{{(x - 2)}}} \right] = - \infty ,{\rm{ do }}\left\{ {\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 1}}{{x + 2}} = \frac{3}{4}}\\{\mathop {\lim }\limits_{x \to {2^ - }} \frac{1}{{x - 2}} = - \infty }\end{array}} \right.}\end{array}\)

d) \(\mathop {\lim }\limits_{x \to - {1^ - }} \frac{{|x + 1|}}{{{x^2} - 1}} = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{ - x - 1}}{{(x - 1)(x + 1)}} = \mathop {\lim }\limits_{x \to - {1^ - }} \frac{{ - 1}}{{x - 1}} = \frac{1}{2}\).