PART 8：例題-連續函數的判斷(05:14)

設 \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{x - 2}}{{\sqrt {x + 2}  - 2}}}&{}\\k&{}\end{array}} \right.\)  \(\begin{array}{l}\left( {x \ne 2} \right)\\\left( {x = 2} \right)\end{array}\) 在 \(x = 2\) 連續，是求 \(k\) 值

SOL:

\(\lim\limits_{x \to 2} \frac{{x - 2}}{{\sqrt {x + 2}  - 2}}\)=  \(\lim\limits_{x \to 2} \frac{{(x - 2)(\sqrt {x + 2}  + 2)}}{{(\sqrt {x + 2}  - 2)(\sqrt {x + 2}  + 2)}}\)=  \(\mathop {\lim }\limits_{x \to 2} \frac{{(x - 2)(\sqrt {x + 2}  + 2)}}{{x + 2 - 4}}\) \(\lim\limits_{x \to 2} \frac{{(x - 2)(\sqrt {x + 2}  + 2)}}{{x - 2}}\) \(\lim\limits_{x \to 2} \sqrt {x + 2}  + 2\) \( = 4\)