A) 0
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: B
Solution :
[b] Let\[f(x)=\frac{x(x-2)}{{{x}^{2}}-4}=\frac{x(x-2)}{(x-2)(x+2)}=\frac{x}{x+2}\] Since f(x) is continuous at x = 2 \[\therefore \underset{x\to 2}{\mathop{\lim }}\,f(x)=f(2)\] \[\Rightarrow \underset{x\to 2}{\mathop{\lim }}\,\frac{x}{x+2}=f(2)\Rightarrow f(2)=\frac{2}{4}=\frac{1}{2}\]You need to login to perform this action.
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