Problem 66 - Calculus: Graphical, Numerical, Algebraic

Question

The Greatest Integer Function (a) Show that $$\frac{x-1}{x}<\frac{\text { int } x}{x} \leq 1(x>0)$$ and $$\frac{x-1}{x}>\frac{\text { int } x}{x} \geq 1(x<0)$$ (b) Determine $$\lim _{x \rightarrow \infty} \frac{\operatorname{int} x}{x}$$ (c) Determine$$\lim _{x \rightarrow-\infty} \frac{\operatorname{int} x}{x}$$

Step-by-Step Solution

Answer

The inequality \( \frac{x-1}{x} < \frac{int(x)}{x} \leq 1 \) holds for all \( x > 0 \), while \( \frac{x-1}{x} > \frac{int(x)}{x} \geq 1 \) holds for all \( x < 0 \). The limits \( \lim _{x \rightarrow \infty} \frac{int(x)}{x} \) and \( \lim _{x \rightarrow-\infty} \frac{int(x)}{x} \) both equal to 1.

Step 1: Proof of first inequality

First, note that \( int(x) \) is the greatest integer less than or equal to \( x \). Therefore, \( x-1 < int(x) \), from which we infer \( \frac{x-1}{x} < \frac{int(x)}{x} \), which proves the left side of the first inequality. For the right side, we observe that \( int(x) \) is always an integer and thus it's clear that \( \frac{int(x)}{x} \) is less than or equal to 1 for all \( x > 0 \). So we get \( \frac{x-1}{x} < \frac{int(x)}{x} \leq 1 \) for all \( x > 0 \).

Step 2: Proof of second inequality

For \( x < 0 \), \( int(x) \) is the greatest integer less than \( x \). It can be inferred that \( x - 1 > int(x) \) which leads to \( \frac{x-1}{x} > \frac{int(x)}{x} \). On the right side, for \( x < 0 \), it's clear that \( \frac{int(x)}{x} \) is greater than or equal to 1 as \( int(x) \) will be less than \( x \). Thus, \( \frac{x-1}{x} > \frac{int(x)}{x} \geq 1 \) for all \( x < 0 \).

Step 3: Calculating the Limit as x approaches Infinity

For \( x > 0 \), \( int(x) \) is at least \( x - 1 \) but it's less than or equal to \( x \). So, \( 1 - \frac{1}{x} = \frac{x-1}{x} \leq \frac{int(x)}{x} \leq 1 \). As \( x \) approaches infinity, both \( \frac{x-1}{x} \) and 1 approach 1. Therefore, by the squeeze theorem, \( \lim _{x \rightarrow \infty} \frac{int(x)}{x} = 1 \).

Step 4: Calculating the Limit as x approaches -Infinity

For \( x < 0 \), \( int(x) \) is at most \( x \) but it's greater than \( x - 1 \). So, \( 1 = \frac{x}{x} \leq \frac{int(x)}{x} < \frac{x-1}{x} = 1 - \frac{1}{x} \). As \( x \) approaches -infinity, both 1 and \( \frac{x-1}{x} \) approach 1. Therefore, by the squeeze theorem, \( \lim _{x \rightarrow-\infty} \frac{int(x)}{x} = 1 \).

Key Concepts

Limits of Functions

Understanding the concept of limits is essential when studying calculus, as it provides insight into a function's behavior as the input approaches a particular value. The limit of a function as the input approaches some value is, in simple terms, the value that the function output 'approaches' as the input gets arbitrarily close to a particular point. For the Greatest Integer Function, we're particularly interested in the limits as x approaches infinity and negative infinity.

When we look at the limit of \( \frac{\text{int } x}{x} \) as x approaches infinity, we're examining the behavior of the integer part of x divided by x itself when x gets larger and larger. Intuitively, as x grows without bound, the difference between x and \( \text{int } x \) (the greatest integer less than or equal to x) becomes negligible. Formally, this concept involves confirming that for all positive values of x, this fraction will eventually stabilize at a specific number as x grows towards infinity.

Inequalities in Calculus

In calculus, inequalities help us to identify the range within which a function's output will lie over its domain. They are particularly useful for establishing bounds on function values and for comparative statics – analyzing how changes in one variable can lead to changes in another.

With the Greatest Integer Function, we applied inequalities to assert that \( \frac{x-1}{x} < \frac{\text{int } x}{x} \leq 1 \) for positive x. Here, the inequality assured us that the function value must lie between two other simpler functions. For negative values of x, a flipped inequality \( \frac{x-1}{x} > \frac{\text{int } x}{x} \geq 1 \) is used due to the properties of the function when x is less than zero. By employing these inequalities, one can also illustrate certain properties of a function graphically, often useful for deeper conceptual understanding.

Squeeze Theorem

The Squeeze Theorem (also known as the Sandwich Theorem) is an invaluable tool in calculus for determining limits that may not be readily apparent. It states that if three functions satisfy the condition that \( f(x) \leq g(x) \leq h(x) \), and if \( \lim_{x \to c} f(x) \) and \( \lim_{x \to c} h(x) \) both exist and are equal to L, then \( \lim_{x \to c} g(x) = L \) as well. It's akin to 'squeezing' g(x) between f(x) and h(x).

In our exercise, we've essentially 'squeezed' our Greatest Integer Function ratio between two other functions as x approaches infinity and negative infinity. Since both bounding functions approach 1, the Squeeze Theorem concludes that the limit of \( \frac{int x}{x} \) must also be 1, both as x approaches positive and negative infinity. Understanding this theorem enriches our ability to solve complex limit problems, especially when direct methods are ineffective or difficult to apply.