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/ has A) 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 \( x=2 \)
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/ has A) 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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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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