Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ + }} \frac{{x - 5}}{{\sqrt {2x - 1} - 3}}\\
= \mathop {\lim }\limits_{x \to {5^ + }} \frac{{\left( {x - 5} \right)\left( {\sqrt {2x - 1} + 3} \right)}}{{\left( {\sqrt {2x - 1} - 3} \right)\left( {\sqrt {2x - 1} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to {5^ + }} \frac{{\left( {x - 5} \right)\left( {\sqrt {2x - 1} + 3} \right)}}{{\left( {2x - 1} \right) - {3^2}}}\\
= \mathop {\lim }\limits_{x \to {5^ + }} \frac{{\left( {x - 5} \right)\left( {\sqrt {2x - 1} + 3} \right)}}{{2\left( {x - 5} \right)}}\\
= \mathop {\lim }\limits_{x \to {5^ + }} \frac{{\sqrt {2x - 1} + 3}}{2}\\
= \frac{{\sqrt {2.5 - 1} + 3}}{2}\\
= \frac{6}{2} = 3\\
\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ - }} \left[ {{{\left( {x - 5} \right)}^2} + 3} \right] = {\left( {5 - 5} \right)^2} + 3 = 3\\
\Rightarrow \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right)
\end{array}\)
Do đó, hàm số đã cho liên tục tại \(x = 5\)
\(\begin{array}{l}
e,\\
\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {1 - \cos x} \right) = 1 - \cos 0 = 0\\
\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \sqrt {x + 1} = \sqrt {0 + 1} = 1\\
\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)
\end{array}\)
Do đó, hàm số đã cho không liên tục tại \(x = 0\)
