Giải thích các bước giải:
\(\begin{array}{l}
b,\\
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {x + 3} - 2} \right)\left( {\sqrt {x + 3} + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 3} \right) - {2^2}}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt {x + 3} + 2}} = \frac{1}{{\sqrt {1 + 3} + 2}} = \frac{1}{4}\\
\Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)
\end{array}\)
Do đó, hàm số đã cho liên tục tại \(x = 1\)
d,
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{{x^2} - 1}}{{\sqrt {2 - x} - 1}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {{x^2} - 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}{{\left( {\sqrt {2 - x} - 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}{{\left( {2 - x} \right) - {1^2}}}\\
= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {\sqrt {2 - x} + 1} \right)}}{{1 - x}}\\
= - \mathop {\lim }\limits_{x \to {1^ - }} \left( {x + 1} \right)\left( {\sqrt {2 - x} + 1} \right)\\
= - \left( {1 + 1} \right).\left( {\sqrt {2 - 1} + 1} \right)\\
= - 4\\
\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( { - 2x} \right) = - 2.1 = - 2\\
\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)
\end{array}\)
Do đó, hàm số đã cho không liên tục tại \(x = 1\)
