Problem 79 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{[x]+[2 x]+[3 x]+\ldots+[n x]}{1+2+3+\ldots+n}=\) (A) \(x\) (B) \(2 x\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

(A) x

Step 1: Understand the Ceiling Function

The expression \([x]\) denotes the greatest integer less than or equal to \(x\). This is the floor function, which will be applied to each term \([k \cdot x]\) for \(k = 1, 2, ..., n\).

Step 2: Simplify the Denominator

The denominator in the limit expression is \(1 + 2 + 3 + \ldots + n\), which is known as the sum of the first \(n\) natural numbers. This sum is given by the formula \(\frac{n(n+1)}{2}\).

Step 3: Analyze the Limit Expression

Rewriting the expression, we have the limit as \(n\) approaches infinity of \[\frac{[x] + [2x] + [3x] + \ldots + [nx]}{\frac{n(n+1)}{2}}.\] These terms in the numerator approximate \(x + 2x + 3x + \ldots + nx - \{x\} - \{2x\} - \ldots - \{nx\}\), where \(\{x\}\) is the fractional part of \(x\).

Step 4: Approximate the Numerator

The expression \([x] + [2x] + [3x] + \ldots + [nx]\) can be approximated by the sum \(x + 2x + 3x + \ldots + nx\) because subtracting the fractional parts contributes a negligible amount as \(n\) grows large.

Step 5: Simplifying the Limit

The approximated sum of the numerator becomes \(x(1 + 2 + 3 + \ldots + n)\), which is equal to \(x \left( \frac{n(n+1)}{2} \right)\). So, the new expression for the limit turns into \(\lim _{n \rightarrow \infty} \frac{x \frac{n(n+1)}{2}}{\frac{n(n+1)}{2}}\).

Step 6: Evaluate the Limit

By simplifying the expression in Step 5, the \(\frac{n(n+1)}{2}\) terms cancel out, leaving us with \(\lim _{n \rightarrow \infty} x\), which gives the value of the limit as \(x\).

Key Concepts

Floor Function

The floor function, often represented as \([x]\), is a mathematical tool used to round down a real number to the nearest integer less than or equal to that number.
For example, \([3.7] = 3\) and \([-2.3] = -3\). The purpose of the floor function is to discard any fractional part of a number and focus only on its integer component.

In the context of this exercise, we apply the floor function to each term \([k \cdot x]\), where \(k\) is a natural number.

• This transforms continuous values into discrete values and simplifies certain calculations.
• The floor function plays a crucial role in understanding how the numerator of our given limit behaves as \(n\) increases towards infinity.

Knowing how to manipulate and understand the floor function helps in simplifying expressions, especially when combining it with other mathematical concepts like limits.

Sum of Natural Numbers

The sum of natural numbers is a fundamental concept in mathematics. When you add up the series of natural numbers from 1 to \(n\), you get the sum \(1 + 2 + \, ...\, + n\).
This sum can be quickly calculated using the formula \[\frac{n(n+1)}{2}\].

In our exercise, this sum forms the denominator of the limit we are examining.

• This sum represents a triangular pattern when visualized graphically, which is why it's sometimes called a triangular number.
• As \(n\) approaches infinity, this sum also grows infinitely large.

Knowing this formula allows us to efficiently handle large sums and simplifies the process of finding the limit. Armed with this knowledge, you can quickly resolve the expression and help unravel the complexities of limits involving arithmetic series.

Fractional Part

The fractional part of a number is what remains after the integer part has been removed.
It is represented using the notation \(\{x\}\), which equals \((x - [x])\). For instance, if \(x = 5.75\), then \([x] = 5\) and \{x\}= 0.75\).

In this exercise, understanding the fractional part helps in approximating the numerator of the expression:

• When applying the floor function to terms \([k\cdot x]\), any fractional part \(\