Problem 43 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

For each \(t \in R\), let \([t]\) be the greatest integer less than or equal to \(\mathrm{t}\). Then \([2018]\) \(\lim _{x \rightarrow 0^{+}} x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{15}{x}\right]\right)\) (a) is equal to 15 . (b) is equal to 120 . (c) does not exist (in \(\mathrm{R}\) ). (d) is equal to 0 .

Step-by-Step Solution

Answer

The limit is equal to 120.

Step 1: Understand the Problem

We are given a fractional part function, where for each number in the form \( t \), \( [t] \) is the greatest integer less than or equal to \( t \). We need to find the limit of a sequence as \( x \) approaches 0 from the positive side. The sequence expression is \( x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{15}{x}\right]\right) \).

Step 2: Analyze the Terms [i/x]

For each \( i = 1, 2, \ldots, 15 \), analyze the term \( \left[\frac{i}{x}\right] \). As \( x \to 0^+ \), \( \frac{i}{x} \to \infty \). Thus, \( \left[\frac{i}{x}\right] \) will increase without bound, approximating to \( \frac{i}{x} \) minus a small fractional part.

Step 3: Rewrite the Expression

Rewrite the primary expression: \[ S(x) = x \sum_{i=1}^{15} \left[ \frac{i}{x} \right] \] As \( x \to 0^+ \), we have: \[ \frac{i}{x} = n_i + \epsilon_i \] where \( n_i \) is an integer and \( \epsilon_i \) is a small fractional part. Thus, \( \left[\frac{i}{x}\right] \approx \frac{i}{x} \).

Step 4: Evaluate the Limit

Replace each term \( \left[\frac{i}{x}\right] \approx \frac{i}{x} \) back into the expression and sum them up:\[ S(x) \approx x \left( \sum_{i=1}^{15} \frac{i}{x} \right) = \sum_{i=1}^{15} i = 120 \]Take the limit as \( x \to 0^+ \):\[ \lim_{x \to 0^+} S(x) = 120 \]

Step 5: Conclusion

The limit evaluates to a constant value, indicating the final answer is option (b), equal to 120.

Key Concepts

Limit of a Function

The limit of a function is a fundamental concept in calculus. It describes the value that a function approaches as the input approaches a specified point. For a function \( f(x) \), the limit \ \( \lim_{x \to a} f(x) \) \ describes what \( f(x) \) approaches as \( x \) gets arbitrarily close to \( a \).
When dealing with limits, it's important to note:

• If \( f(x) \) gets closer to a specific number \( L \) as \( x \) approaches \( a \), then \( \lim_{x \to a} f(x) = L \).
• In some cases, a function may not have a limit as \( x \) approaches \( a \), which can happen if the function behaves erratically.

This particular exercise considers the limit as \( x \) approaches 0 from the positive side, denoted as \( x \to 0^+ \). This means that \( x \) comes from the right side of zero, gradually decreasing towards zero but always remaining positive. Understanding the direction from which \( x \) approaches a number can be crucial in finding the correct limit.

Convergence

Convergence is the idea that a sequence or series approaches a certain value as the index increases indefinitely. In the exercise, the term \( x[\frac{i}{x}] \) is evaluated, and we need to assess whether this expression converges to a certain number as \( x \) approaches zero.

• When a sequence converges, it gets closer and closer to a particular value.
• Convergence is crucial in calculus as it helps in determining the stability and behavior of sequences and series.

For the given sequence \( x\left([\frac{1}{x}]+[\frac{2}{x}]+\ldots+[\frac{15}{x}]\right) \), as \( x \to 0^+ \), the terms \([\frac{i}{x}]\) become very large, while the multiplication by \( x \) brings the function towards a stable value, ultimately leading to convergence towards the result 120.

Sequence and Series

A sequence is an ordered list of numbers, following a particular pattern or rule. A series is essentially the sum of the terms of a sequence. In the problem, we deal with a series formed by summing specific sequence terms \[ \left[\frac{1}{x}\right], \left[\frac{2}{x}\right], \ldots, \left[\frac{15}{x}\right] \].

• The notation \( [t] \) represents the greatest integer function, providing the greatest integer less than or equal to \( t \).
• Summing the results of this greatest integer function, as \( x \) decreases toward zero, forms the basis of a series in the problem.

To find the behavior of this series as \( x \to 0^+ \), it is crucial to understand how each term contributes and how they combine to form the total. This combination is evaluated in Step 4 of the solution, showing that the series stabilizes to a sum of 120 as \( x \) approaches zero.

Fractional Part Function

The fractional part function finds the fractional portion of a number, often represented as \( \{x\} \). This is calculated as the original number minus its greatest integer part. When working with expressions like \( \frac{i}{x} \), being aware of the fractional part helps explain why \( \left[\frac{i}{x}\right] \approx \frac{i}{x} \).

• The expression \( \left[\frac{i}{x}\right] \) removes the fractional part, rounding down to the nearest whole number less than or equal to \( \frac{i}{x} \).
• As \( x \) approaches zero, \( \frac{i}{x} \) increases, leading to smaller fractional parts relative to the entire term.

Understanding the fractional part is key when breaking down complex sequences and series into manageable parts. In this exercise, the approximation used for \( \left[\frac{i}{x}\right] \) stems from this fractional part function, facilitating easier evaluation of the limit.