If mathop {lim }limits_{n to infty } frac{1 - | vedclass

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

Vedclass · Accepted Answer

$1$

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(C) Given the limit: $\mathop {\lim }\limits_{n 	o \infty } \frac{1 - 10^n}{1 + 10^{n+1}}$Divide the numerator and denominator by $10^n$:$= \mathop {\lim }\limits_{n 	o \infty } \frac{\frac{1}{10^n} - 1}{\frac{1}{10^n} + 10}$As $n 	o \infty$,$\frac{1}{10^n} 	o 0$:$= \frac{0 - 1}{0 + 10} = -\frac{1}{10}$Comparing this with the given expression $\frac{-\alpha}{10}$,we get:$-\frac{1}{10} = -\frac{\alpha}{10}$Therefore,$\alpha = 1$.

If $\mathop {\lim }\limits_{n \to \infty } \frac{1 - 10^n}{1 + 10^{n+1}} = \frac{-\alpha}{10}$,then find the value of $\alpha$.

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