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If $\lim _{n \rightarrow \infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\frac{-\alpha}{10}$, then $\alpha$ is equal to
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$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{1-10^n}{1+10^{n+1}}=-\frac{\alpha}{10} \\ \Rightarrow & \lim _{n \rightarrow \infty} \frac{1-10^n}{1+10 \cdot 10^n}=-\frac{\alpha}{10} \\ & \lim _{n \rightarrow \infty} \frac{10^n\left[\frac{1}{10^n}-1\right]}{10^n\left[\frac{1}{10^n}+10\right]}=-\frac{\alpha}{10}\end{aligned}$
$\begin{aligned} & \because n \rightarrow \infty, 10^n \rightarrow \infty, \frac{1}{10^n} \rightarrow 0 \\ & \Rightarrow \quad \frac{0-1}{0+10}=\frac{-\alpha}{10} \Rightarrow \frac{-1}{10}=-\frac{\alpha}{10} \\ & \therefore \quad \alpha=1\end{aligned}$
$\begin{aligned} & \because n \rightarrow \infty, 10^n \rightarrow \infty, \frac{1}{10^n} \rightarrow 0 \\ & \Rightarrow \quad \frac{0-1}{0+10}=\frac{-\alpha}{10} \Rightarrow \frac{-1}{10}=-\frac{\alpha}{10} \\ & \therefore \quad \alpha=1\end{aligned}$
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