Functions Question 317
Question: $ \underset{n\to \infty }{\mathop{\lim }},\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+…+\frac{1}{2^{n}} $ equals
[RPET 1996]
Options:
A) 2
B) ?1
C) 1
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\underset{n\to \infty }{\mathop{\lim }}\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+…….+\frac{1}{2^{n}}=\underset{n\to \infty }{\mathop{\lim }},\frac{1}{2},\frac{[ 1-{{( \frac{1}{2} )}^{n}} ]}{( 1-\frac{1}{2} )} $ $ \underset{n\to \infty }{\mathop{\lim }}[ 1-\frac{1}{2^{n}} ]=1-0=1 $
BETA
