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Prove that \(f(x) = \begin{cases} \frac {x^2-25}{x}, & \quad \text{when x≠5;} \text{}\\ 10, & \quad \text{when x=5} \end{cases}\) is continuous at x = 5 |
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Answer» LHL: = \(\lim\limits_{x \to5^-} \)f(x) = \(\lim\limits_{x \to5^-} \) \(\frac{x^2 - 25}{x-5} \) = \(\lim\limits_{x \to5^-} \) \(\frac{(x+5)(x-5)}{x-5} \)[By middle term splitting] = \(\lim\limits_{x \to5^-} \) x + 5 = 10 RHL: = \(\lim\limits_{x \to5^+} \)f(x) = \(\lim\limits_{x \to5^+} \) \(\frac{x^2 - 25}{x-5} \) = \(\lim\limits_{x \to5^+} \) \(\frac{(x+5)(x-5)}{x-5} \)[By middle term splitting] = \(\lim\limits_{x \to5^+} \) x + 5 = 10 f(5)= 10 Since, = \(\lim\limits_{x \to5} \) f(x) = f(5) f is continuous at x=5. |
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