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Previous Year Question (PYQs)

For $x \in \mathbb{R}$, find $\displaystyle \lim_{x \to \infty} \left(\dfrac{x - 3}{x + 2}\right)^x = ?$




Solution

Let $L = \left(\dfrac{x - 3}{x + 2}\right)^x = \left(1 - \dfrac{5}{x + 2}\right)^x.$ Take $\ln L = x \ln\!\left(1 - \dfrac{5}{x + 2}\right).$ For large $x$, use $\ln(1 - y) \approx -y$. So, $\ln L \approx x \left(-\dfrac{5}{x + 2}\right) \to -5.$ Hence, $L = e^{-5}.$

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