COMEDK · Maths · 16. Limits
The value of \(\lim _{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{2 x}\) is equal to
- A \(\frac{a+b}{2}\)
- B \(\frac{a-b}{2}\)
- C \(\frac{e^{a b}}{2}\)
- D 0
Answer & Solution
Correct Answer
(B) \(\frac{a-b}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{2 x} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+a x+\frac{(a x)^2}{2 !}+\frac{(a x)^3}{3 !}+\ldots\right)}{-\left(1+b x+\frac{(b x)^2}{2 !}+\frac{(b x)^3}{3 !}\right)} \\ & =\lim _{x \rightarrow 0}…
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