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MHT CET · Maths · Limits

\(\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=\)

  1. A \(\frac{-\mathrm{ab}}{2} \log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\)
  2. B \(\frac{\mathrm{ab}}{2} \log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\)
  3. C \(a b \log \left(\frac{b}{a}\right)\)
  4. D \(-a b \log \left(\frac{b}{a}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\mathrm{ab}}{2} \log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\)

Step-by-step Solution

Detailed explanation

Let \(\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=L\)
\(
\begin{aligned}
& \Rightarrow L=\lim _{x \rightarrow 1} \frac{\left(a b^x \log b\right)-\left(a^x \log a \cdot b\right)}{2 x} \\
& =\frac{a b \log b-a b \log a}{2}=\frac{a b}{2} \log \left(\frac{b}{a}\right)
\end{aligned}
\)
... [L' Hospital rule]