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Prove that \(f(x) = \begin{cases} \frac {x^2-x-6}{x-3}, & \quad \text{when x≠3;} \text{}\\ 5, & \quad \text{when x=3} \end{cases}\) is continuous at x = 3 5, when x = 3 |
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Answer» LHL: \(\lim\limits_{x \to3^-} \)f(x) = \(\lim\limits_{x \to3^-} \) \(\frac{x^2-x-6}{x-3}\) = \(\lim\limits_{x \to3^-} \) \(\frac{(x+2)(x-3)}{x-3}\) [By middle term splitting] = \(\lim\limits_{x \to3^-} \) x + 2 = 5 RHL: \(\lim\limits_{x \to3^-} \)f(x) = \(\lim\limits_{x \to3^-} \) \(\frac{x^2-x-6}{x-3}\) = \(\lim\limits_{x \to3^-} \) \(\frac{(x+2)(x-3)}{x-3}\) [By middle term splitting] = \(\lim\limits_{x \to3^-} \) x + 2 = 5 f(3) = 5 Since, \(\lim\limits_{x \to3} \) f(x) = f(3) f is continuous at x=3. |
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