MCQ
જો $\mathop {\lim }\limits_{n \to \infty } \frac{{1 - {{(10)}^n}}}{{1 + {{(10)}^{n + 1}}}} = \frac{{ - \alpha }}{{10}}$, તો $\alpha = . . .$
- A$0$
- B$-1$
- ✓$1$
- D$2$
$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{{(10)}^n}\left[ {{{\left( {\frac{1}{{10}}} \right)}^n} - 1} \right]}}{{{{(10)}^{n + 1}}\left( {1 + \frac{1}{{{{10}^{n + 1}}}}} \right)}} = - \frac{1}{{10}}$
$\therefore \alpha = 1$.
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