If lim _{n rightarrow infty} frac{1-(10)^n}{1 | vedclass

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(C) Given the limit: $\lim _{n \rightarrow \infty} \frac{1-10^n}{1+10^{n+1}}=-\frac{\alpha}{10}$Divide the numerator and denominator by $10^n$:$\lim _

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$1$

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(C) Given the limit: $\lim _{n \rightarrow \infty} \frac{1-10^n}{1+10^{n+1}}=-\frac{\alpha}{10}$Divide the numerator and denominator by $10^n$:$\lim _{n \rightarrow \infty} \frac{\frac{1}{10^n}-1}{\frac{1}{10^n}+10}=-\frac{\alpha}{10}$As $n \rightarrow \infty$,$\frac{1}{10^n} \rightarrow 0$.Substituting this value,we get:$\frac{0-1}{0+10} = -\frac{\alpha}{10}$$-\frac{1}{10} = -\frac{\alpha}{10}$Therefore,$\alpha = 1$.

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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