อ้างอิง:
ข้อความเดิมเขียนโดยคุณ tonklaZolo
$\lim_{n \to \infty} (1-\frac{1}{8n})^n=?$
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Let $L = (1-\frac{1}{8n})^n$
Take $ln$ into both sides
$ln(L) = ln(1-\frac{1}{8n})^n = n (ln(1-\frac{1}{8n})) = \frac{ln(1-\frac{1}{8n})}{\frac{1}{n}}$
Take Limit $n \rightarrow \infty$
$\lim_{n \to \infty}(ln(L)) = \lim_{n \to \infty}(\frac{ln(1-\frac{1}{8n})}{\frac{1}{n}})$ $ (I.F. \frac{0}{0})$
Use L'Hospital's rule, we have
$ln(\lim_{n \to \infty}(L)) = \lim_{n \to \infty}(\frac{\frac{1}{1-\frac{1}{8n}}\cdot \frac{1}{8n^2}}{\frac{-1}{n^2}}) = \lim_{n \to \infty}(\frac{1}{8(\frac{1}{8n}-1)}) = \lim_{n \to \infty}(\frac{1}{\frac{1}{n}-8}) = \frac{-1}{8}$
$\lim_{n \to \infty}(L) = e^{\frac{-1}{8}} \approx 0.8825$
Hence $\lim_{n \to \infty} (1-\frac{1}{8n})^n = 0.8825 $