Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to 1} \frac{{2x - 1}}{{{{\left( {x - 1} \right)}^2}}}\\
\mathop {\lim }\limits_{x \to 1} \left( {2x - 1} \right) = 2.1 - 1 = 1\\
\mathop {\lim }\limits_{x \to 1} {\left( {x - 1} \right)^2} = {\left( {1 - 1} \right)^2} = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{2x - 1}}{{{{\left( {x - 1} \right)}^2}}}\left( { = \frac{1}{0}} \right) = + \infty \\
b,\\
x \to - \infty \Rightarrow x < - 1 \Rightarrow \sqrt {{x^2} - 1} = \sqrt {1 - x} .\sqrt { - 1 - x} \\
\mathop {\lim }\limits_{x \to - \infty } \frac{{x - 1}}{{\sqrt {{x^2} - 1} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - {{\left( {\sqrt {1 - x} } \right)}^2}}}{{\sqrt {1 - x} .\sqrt { - 1 - x} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 - x} }}{{\sqrt { - 1 - x} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {\frac{{ - 1}}{x} + 1} }}{{\sqrt {\frac{1}{x} + 1} }}\\
= \frac{{ - \sqrt 1 }}{{\sqrt 1 }}\\
= - 1\\
c,\\
\mathop {\lim }\limits_{x \to 1} \frac{{x + 1 - \sqrt {x + 3} }}{{{x^2} - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \left( {2 - \sqrt {x + 3} } \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \frac{{4 - x - 3}}{{2 + \sqrt {x + 3} }}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{1 - \frac{1}{{2 + \sqrt {x + 3} }}}}{{x + 1}}\\
= \frac{{1 - \frac{1}{{2 + \sqrt {1 + 3} }}}}{{1 + 1}}\\
= \frac{{1 - \frac{1}{4}}}{2} = \frac{3}{8}\\
d,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - x} - 1}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{1 - x - 1}}{{\left( {\sqrt {1 - x} + 1} \right)x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{\left( {\sqrt {1 - x} + 1} \right)x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{\sqrt {1 - x} + 1}}\\
= \frac{{ - 1}}{{\sqrt {1 - 0} + 1}}\\
= \frac{{ - 1}}{2}
\end{array}\)
