Problem 78 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

\(\lim _{x \rightarrow 1} 1-x+[x-1]+[1-x] .\)

Step-by-Step Solution

Answer

The given expression is \( 1-x+[x-1]+[1-x] \), where [x] represents the greatest integer function. After analyzing the behavior of the floor function terms near 1 and simplifying the expression, we obtain -x for both intervals (\(0.9 < x <1\)) and (\(1 < x < 1.1\)). Taking the limit as x approaches 1, we get \(\lim _{x \rightarrow 1} -x = -1\). Therefore, the final answer is \(\lim _{x \rightarrow 1} 1-x+[x-1]+[1-x] = -1\).

Step 1: Understand the floor function

The floor function [x] returns the largest integer less than or equal to x. For example: \[ [1.2] = 1, [-1.8] = -2 \] We need to understand the behavior of [x-1] and [1-x] as x approaches 1.

Step 2: Analyze the behavior of the floor function terms near 1

As x approaches 1, consider the two intervals: \( 0.9 < x < 1 \) and \( 1 < x < 1.1\). For these intervals, analyze the values of [x-1] and [1-x]. Case 1: When \(0.9 < x < 1\), \[ x - 1 < 0 \Rightarrow [x-1] = -1 \] \[ 1 - x > 0 \Rightarrow [1-x] = 0 \] Case 2: When \(1 < x < 1.1\), \[ x - 1 > 0 \Rightarrow [x-1] = 0 \] \[ 1 - x < 0 \Rightarrow [1-x] = -1 \]

Step 3: Substitute the values of the floor function terms

Now, substitute the values of [x-1] and [1-x] in the original expression and analyze the limit. Case 1: When \(0.9 < x < 1\), \[ 1 - x + [x-1] + [1-x] = 1 - x - 1 + 0 = -x \] Case 2: When \(1 < x < 1.1\), \[ 1 - x + [x-1] + [1-x] = 1 - x + 0 - 1 = -x \] The expression simplifies to -x in both cases.

Step 4: Take the limit as x approaches 1

As the expression simplifies to -x, we can take the limit as x approaches 1: \[ \lim _{x \rightarrow 1} -x = -1 \] So, the final answer is: \[ \lim _{x \rightarrow 1} 1-x+[x-1]+[1-x] = -1 \]

Key Concepts

Understanding the Floor Function

The floor function, denoted by \([x]\), plays a crucial role in mathematics. It returns the largest integer less than or equal to a given number. For example:

• \([1.2] = 1\)
• \([-1.8] = -2\)

In our problem, we need to consider the behavior of \([x-1]\) and \([1-x]\) as \(x\) approaches 1. This involves examining how the floor function handles values near an integer. Since the floor function captures abrupt changes, it's essential to understand what happens as \(x\) translates across specific intervals. For instance, if \(x\) is slightly less than 1, \([x-1]\) becomes -1 and \([1-x]\) is 0. This is because:

• When \(x < 1\), \(x-1\) is negative, leading \([x-1]\) to drop to the next lower integer, which is -1.
• When \(x > 1\), \([x-1]\) equals 0, since \(x-1\) becomes positive, landing within the integer range directly.

These properties allow us to simplify complex expressions and handle unpredictability in calculations.

The Role of Calculus in Understanding Limits

Calculus is vital in exploring mathematical concepts like limits. A limit helps us understand the behavior of functions as they approach a specific point. By analyzing limits, we can predict the value that a function approaches as the input gets closer to a certain number.In our problem, we need to find \(\lim _{x \rightarrow 1} 1-x+[x-1]+[1-x]\). Here, calculus helps us determine the value of an expression as \(x\) closely approaches 1. The limit captures the essence of the function's behavior as it nears a specific point, even if the exact point is undefined in simple terms.Analyzing limits involves several steps:

• Understand how the values change in the given range.
• Examine any abrupt alterations due to functions like the floor function.
• Simplify the expression at different intervals to ensure clarity.

In the exercise, whether \(x\) approaches from below or above 1, the expression eventually simplifies to \(-x\), leading the limit to be -1.

Interval Analysis to Simplify Calculations

Interval analysis involves dividing the domain around a point of interest into sections or intervals. This technique allows us to address different scenarios distinctly and helps in understanding fluctuating behaviors in functions.For our problem, we considered two intervals around \(x=1\):

• \(0.9 < x < 1\)
• \(1 < x < 1.1\)

In each interval, examining \([x-1]\) and \([1-x]\) gives us different integer outputs. These calculations are:

• In \(0.9 < x < 1\): \([x-1] = -1\) and \([1-x] = 0\)
• In \(1 < x < 1.1\): \([x-1] = 0\) and \([1-x] = -1\)

By analyzing each interval separately, we simplify the expression for each scenario. This technique ensures that complex behaviors are captured accurately, allowing clear insight into the function as \(x\) approaches 1.