Differentiation Question 158
Question: If [.] denotes the greatest integer function, then $ \underset{n\to \infty }{\mathop{\lim }}\frac{[x]+[2x]+…+[nx]}{n^{2}} $ is
Options:
A) 0
B) $ x $
C) $ \frac{x}{2} $
D) $ \frac{x^{2}}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ nx-1<| nx |\le nx. $ Putting $ n=1,2,3,…,n $ and adding them, $ x\Sigma n-n<\sum [nx]\le x\Sigma n $
$ \therefore x.\frac{\Sigma n}{n^{2}}-\frac{1}{n}<\frac{\Sigma [nx]}{n^{2}}\le x.\frac{\Sigma n}{n^{2}} $
- (i) Now, $ \underset{x\to \infty }{\mathop{\lim }}{ x.\frac{\Sigma n}{n^{2}}-\frac{1}{n} }=x.\underset{n\to \infty }{\mathop{\lim }}\frac{\Sigma n}{n^{2}}-\underset{n\to \infty }{\mathop{\lim }}\frac{1}{n}=\frac{x}{2} $ As the two limits are equal, by (i) $ \underset{n\to \infty }{\mathop{\lim }}\frac{\Sigma [nx]}{n^{2}}=\frac{x}{2}. $
BETA
