JEE Main On 08 April 2018 Question 20
Question: For each $ t\in \mathbf{R}, $ let[t] be the greatest integer less than or equal to t. Then $ \underset{x\to 0+}{\mathop{\lim }},x( [ \frac{1}{x} ]+[ \frac{2}{x} ]+……+[ \frac{15}{x} ] ) $ [JEE Main Online 08-04-2018]
Options:
A) is equal to 120.
B) does not exist (in R).
C) is equal to 0.
D) is equal to 15.
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to {0^{+}}}{\mathop{\lim }},x[ \frac{1}{x}-{ \frac{1}{x} }+\frac{2}{x}-{ \frac{2}{x} }+…..+\frac{15}{x}-{ \frac{15}{x} } ] $
$ =120-\underset{x\to {0^{+}}}{\mathop{\lim }},x\underbrace{[ { \frac{1}{x} }+{ \frac{2}{x} }+…..+{ \frac{15}{x} } ]} _{finite} $ $ =120-0=120 $ .
