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KCET · Maths · Limits

If \(\lim\limits_{x \to 3}\left(\dfrac{x^2 - ax - 3b}{x - 3}\right) = 5\), then \(a + b = \)

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

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

For the limit to exist and be finite, the expression must take the \(\dfrac{0}{0}\) form at \(x = 3\).

Substituting \(x = 3\) in the numerator gives:

\(3^2 - a(3) - 3b = 0\)

\(9 - 3a - 3b = 0\)

\(3(a + b) = 9\)

\(a + b = 3\)

Applying L'Hopital's rule to find the individual values:

\(\lim\limits_{x \to 3} \dfrac{2x - a}{1} = 5\)

\(6 - a = 5 \Rightarrow a = 1\)

Substituting \(a = 1\) into \(a + b = 3\) gives \(b = 2\).

Thus, \(a + b = 1 + 2 = 3\).

Answer: \(3\)