PART 5：例題-函數不連續的判斷(09:42)

下列函數在何處不連續?

1. \(f\left( x \right) = \frac{{{x^2} - 1}}{{x + 1}}\)

2. \(f\left( x \right) = \frac{{x + 7}}{{{x^2} - 2x - 3}}\)

3. \(f\left( x \right) = \frac{x}{{\left| x \right| - 3}}\)

4. \(f\left( x \right) = \frac{{x + 3}}{{\left| {{x^2} + 3x} \right|}}\)

5. \(f\left( x \right) = \) \(\left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2} - 1}}{{x + 1}}}&{}\\6&{}\end{array}} \right.\) \(\begin{align}
& \left( 若x\ne -1 \right) \\
& \left( 若x=-1 \right) \\
\end{align}\)
SOL:
1. \(f\left( x \right) = \frac{{{x^2} - 1}}{{x + 1}}\)  \( \Rightarrow \;\)  \(x =  - 1\) ，

2.  \(f\left( x \right) = \frac{{x + 7}}{{{x^2} - 2x - 3}} = \frac{{x + 7}}{{(x - 3)(x + 1)}}\) \( \Rightarrow \;\)  \(x =  - 1\) ， \(x = 3\)

3.  \(f\left( x \right) = \frac{x}{{\left| x \right| - 3}}\)  \( \Rightarrow \left| x \right| - 3 = 0\) ， \(x =  - 3\;\;\) ， \(x = 3\)

4. \(f\left( x \right) = \frac{{x + 3}}{{\left| {{x^2} + 3x} \right|}} = \frac{{x + 3}}{{\left| x \right|\left| {x + 3} \right|}}\)   \( \Rightarrow x = 0\) ， \(x =  - 3\;\)

5. \(f\left( x \right) = \) \(\left\{ {\begin{array}{*{20}{c}}{\frac{{{x^2} - 1}}{{x + 1}}}&{}\\6&{}\end{array}} \right.\) \(\begin{align}
& \left( 若x\ne -1 \right) \\
& \left( 若x=-1 \right) \\
\end{align}\)， \( \Rightarrow \) \(\lim\limits_{x \to  - 1} \frac{{{x^2} - 1}}{{x + 1}}\)  \( \Rightarrow \)  \(\lim\limits_{x \to  - 1} \frac{{(x + 1)(x - 1)}}{{x + 1}} =  - 2 \ne 6\)  \( \Rightarrow x =  - 1\)