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If the function \(f(x)=\begin{cases}\cfrac{x^2-1}{x-1} & \quad \text when\,x \neq 1\\k & \quad \text when\,x =1\end{cases}\), is given to be continuous at x = 1, then what is the value of k ? |
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Answer» Left hand limit of function\(f\) at \(x\) = 1 is \(f\)(1 −) = \(\lim\limits_{x \to1} f(x)=\lim\limits_{x \to 1}\cfrac{x^2-1}{x-1} \) \(=\lim\limits_{x \to 1}\cfrac{(x-1)(x+1)}{x-1} \) = \(\lim\limits_{x \to1}\,x+ 1\) = 1 + 1 = 2. Hence, Left hand limit of function\(f\) at \(x\) = 1 is \(f\) (1 −) = 2. Since, given that the function \(f\)(\(x\)) is continuous at \(x\) = 1. Therefore, \(f\)(1 −) = \(f\)(1) = \(f\)(1 +), where \(f\)(1 +) is right hand limit of function \(f\)(\(x\)) at \(x\) = 1. Now, \(f\)(1 −) = \(f\)(1) ⇒ k = 2. ( \(\because\) \(f\)(1) = and \(f\)(1 −) = 2 ) Hence, if k = 2, then the function \(f\) (\(x\)) is continuous at \(x\) = 1. |
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