A) 0
B) ?1
C) 1
D) 2
Correct Answer: C
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1-{{(10)}^{n}}}{1+{{(10)}^{n+1}}}\]\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(10)}^{n}}\left[ {{\left( \frac{1}{10} \right)}^{n}}-1 \right]}{{{(10)}^{n+1}}\left( 1+\frac{1}{{{10}^{n+1}}} \right)}=-\frac{1}{10}\] \[\therefore \alpha =1\].
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