Differentiation Question 131
Question: Let $ f(x)=x-[x], $ where [x] denotes the greatest integer $ \le x $ and $ g(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{{{{f(x)}}^{2n}}-1}{{{{f(x)}}^{2n}}+1}, $ then g(x) is equal to
Options:
A) 0
B) 1
C) -1
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] As $ 0\le x-[x]<1\forall x\in R,0\le f(x)<1 $
$ \therefore \underset{n\to \infty }{\mathop{\lim }}{{{f(x)}}^{2n}}=0 $ Thus, for $ x\in R, $
$ g(x)=\underset{n\to \infty }{\mathop{\lim }}\frac{{{{f(x)}}^{2n}}-1}{{{{f(x)}}^{2n}}+1}=\frac{0-1}{0+1}=-1. $
BETA
