Question:medium
If \(\lim_{x \to 3} \frac{(2x - k)\tan(x - 3)}{x^2 - 6x + 9} = 2\), then the value of 'k' is equal to
If \(\lim_{x \to 3} \frac{(2x - k)\tan(x - 3)}{x^2 - 6x + 9} = 2\), then the value of 'k' is equal to
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For a limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) to exist and be finite when \(g(c) = 0\), \(f(c)\) must also be 0. Setting the numerator to zero at the limit point is a common trick to find unknown constants.
Updated On: Jun 25, 2026
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The Correct Option is D
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