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4. \( \quad \begin{aligned}f(x)\} &=\frac{x^{2}-4}{x-2} \quad x \neq 2 \\ &=2 \quad x=2 \end{aligned} \)Then \( f(x) \) is/ hasA) Continuous at \( x=2 \)B) Removable discontinuity at \( x=2 \)C) Jump discontinuity at \( x=2 \)D) Infinite discontinuity at \( x=2 \)E) missing point discontinuity at \( x=2 \) |
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Answer» f(x) = \(\begin{cases} \frac{x^2-4}{x-2}, & \quad x\neq2\\ 2, & \quad x = 2 \end{cases}\) \(\lim\limits_{x\to2}f(x)=\lim\limits_{x\to2}\frac{x^2-4}{x-2}\) \(=\lim\limits_{x\to2}\frac{(x-2)(x+2)}{x-2}\) \(=\lim\limits_{x+2}\,x+2\) = 2 + 2 = 4 But f(2) = 2 Therefore \(\lim\limits_{x\to 2}f(x)\neq f(2)\) Thus, f(x) has a removable discontinuity. We have to change the value of function at x = 2 which is 4. |
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