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COMEDK · Maths · 16. Limits

The value of \(\lim _\limits{x \rightarrow 0} \dfrac{e^{a x}-e^{b x}}{2 x}\) is equal to

  1. A 0
  2. B \(\dfrac{e^{a b}}{2}\)
  3. C \(\dfrac{a+b}{2}\)
  4. D \(\dfrac{a-b}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\dfrac{a-b}{2}\)

Step-by-step Solution

Detailed explanation

The limit is given by \(L = \lim_{x \rightarrow 0} \dfrac{e^{ax} - e^{bx}}{2x}\). Since the limit is of the form \(\dfrac{0}{0}\) as \(x \rightarrow 0\), L'Hopital's rule can be applied by differentiating the numerator and the denominator with respect to \(x\).…