$\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=$
$\lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=$
- $\frac{-\mathrm{ab}}{2} \log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)$
- $\frac{\mathrm{ab}}{2} \log \left(\frac{\mathrm{b}}{\mathrm{a}}\right)$
- $a b \log \left(\frac{b}{a}\right)$
- $-a b \log \left(\frac{b}{a}\right)$
Solution
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]
Asked in: MHT CET 2021 (23 Sep Shift 1)
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