Functions Question 298
Question: If $ \underset{n\to \infty }{\mathop{\lim }},\frac{1-{{(10)}^{n}}}{1+{{(10)}^{n+1}}}=\frac{-\alpha }{10} $ , then give the value of $ \alpha $ is
[Orissa JEE 2005]
Options:
A) 0
B) ?1
C) 1
D) 2
Show Answer
Answer:
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}}[ {{( \frac{1}{10} )}^{n}}-1 ]}{{{(10)}^{n+1}}( 1+\frac{1}{{10^{n+1}}} )}=-\frac{1}{10} $
$ \therefore \alpha =1 $ .
BETA
