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JEE Mains · Maths · STD 11 - 12. limits

If \(\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5\) , then \(k\) is equal to

  1. A \(0\)
  2. B \(1\)
  3. C \(2\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3\)

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Detailed explanation

\(\left( d \right)\,\,\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\left\{ {{x^2} + \left( {k - 2} \right)x - 2k} \right\}}}{{{x^2} - 4x + 4}} = 5\)…
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