Problem 54 - Calculus: Early Transcendentals

Question

Let \( f(x) = [ \cos x ] \), \( -\pi \le x \le \pi \). (a) Sketch the graph of \( f \). (b) Evaluate each limit, if it exists. (i) \( \displaystyle \lim_{x \to 0}f(x) \) (ii) \( \displaystyle \lim_{x \to (\pi/2)^-}f(x) \) (iii) \( \displaystyle \lim_{x \to (\pi/2)^+}f(x) \) (iv) \( \displaystyle \lim_{x \to \pi/2}f(x) \) (c) For what values of \( a \) does \( \displaystyle \lim_{x \to a}f(x) \) exist?

Step-by-Step Solution

Answer

The limit is undefined at odd multiples of \(\frac{\pi}{2}\); defined elsewhere.

Step 1: Analyze the Function

The function \( f(x) = [ \cos x ] \) is a greatest integer function (also known as the floor function) that takes each value of \( \cos x \) to the greatest integer less than or equal to it. The cosine function oscillates between -1 and 1 on the given interval \( -\pi \le x \le \pi \). Therefore, \( f(x) \) takes values from this interval to the nearest lower integer, typically staying constant within certain ranges, as the cosine function varies smoothly.

Step 2: Sketch the Graph of f(x)

The graph of \( f(x) \) will appear as a series of step functions. Since \( \cos x = 1 \) at \( x=0 \) and decreases to \( \cos x = 0 \) at \( x = \frac{\pi}{2} \), and \( \cos x = -1 \) at \( x = \pi \), the value stays at 1, then steps down to 0 at \( x = \frac{\pi}{2} \), and steps down to -1 at \( x = \pi \). This pattern repeats symmetrically about the y-axis (since cosine is an even function).

Step 3: Evaluate the Limit at x = 0

For \( \lim_{x \to 0} f(x) \), since \( \cos(0) = 1 \), the function \( f(x) = 1 \) near \( x=0 \). Thus, \( \lim_{x \to 0} f(x) = 1 \).

Step 4: Evaluate the Left-Hand Limit at \( x = \frac{\pi}{2} \)

For \( \lim_{x \to (\frac{\pi}{2})^-} f(x) \), the values of \( \cos x \) just to the left of \( x = \frac{\pi}{2} \) approach 0 from above. Because \( f(x) \) is the floor function, just before \( x = \frac{\pi}{2} \), \( f(x) \) would be 0. Thus, \( \lim_{x \to (\frac{\pi}{2})^-} f(x) = 0 \).

Step 5: Evaluate the Right-Hand Limit at \( x = \frac{\pi}{2} \)

For \( \lim_{x \to (\frac{\pi}{2})^+} f(x) \), right after \( x = \frac{\pi}{2} \), the cosine values are slightly negative. The greatest integer less than or equal to these would be -1, so \( f(x) = -1 \). Thus, \( \lim_{x \to (\frac{\pi}{2})^+} f(x) = -1 \).

Step 6: Evaluate the Limit at \( x = \frac{\pi}{2} \)

Since \( \lim_{x \to (\frac{\pi}{2})^-} f(x) = 0 \) and \( \lim_{x \to (\frac{\pi}{2})^+} f(x) = -1 \), the limit \( \lim_{x \to \frac{\pi}{2}} f(x) \) does not exist due to the discontinuity at \( x = \frac{\pi}{2} \).

Step 7: Determine Where the Limit Exists

\( f(x) = [ \cos x ] \) is discontinuous at points where \( \cos x \) passes an integer value, as it causes a step change in function value. However, for continuous intervals where \( \cos x eq n \) (\( n \) being an integer), the function remains constant, so the limit exists at all other points, particularly where \( cos(x) eq 0 \). Therefore, \( \lim_{x \to a} f(x) \) exists at all \( a \) other than odd multiples of \( \frac{\pi}{2} \).

Key Concepts

Graph Sketching

Graph sketching is essential in visualizing and understanding functions. For sketching the graph of the function \( f(x) = [ \cos x ] \), note that it represents the floor of the cosine of \( x \). This means the output is the greatest integer less than or equal to \( \cos(x) \). The cosine function oscillates smoothly between -1 and 1, but the floor function transforms these into a step-like graph.
This graph will have steps at the points where cosine values are equal to whole numbers. For \( x = 0 \), \( \cos(0) = 1 \), so \( f(x) = 1 \). As \( x \) approaches \( \pi/2 \) from the left, \( \cos(x) \) reaches just above 0, making \( f(x) = 0 \). Crossing \( x = \pi/2 \) to \( \pi \), \( \cos(x) \) dips negative, shifting \( f(x) \) to be -1. Thus, the graph is a series of horizontal lines with jumps or 'steps' at \( x = \pm \pi/2 \).
Graphically, it'll appear as a step function which fully mimics the cosines changes only in jumps and not in continuous tendencies.

Limit Evaluation

Limit evaluation involves determining the value a function approaches as the input approaches a certain point. It’s essential to understand the behavior of \( f(x) = [ \cos x ] \) at key points. Let's consider the following limits:

• For \( \lim_{x \to 0} f(x) \), since \( \cos(0) = 1 \), the function approaches 1 near 0. Therefore, \( \lim_{x \to 0} f(x) = 1 \).
• For \( \lim_{x \to (\pi/2)^-} f(x) \), \( \cos(x) \) approaches 0 from above, resulting in \( f(x) = 0 \) just before \( x = \pi/2 \).
• For \( \lim_{x \to (\pi/2)^+} f(x) \), immediately after \( x = \pi/2 \), \( \cos(x) \) values slightly negative. Thus, \( f(x) = -1 \).
• Because these left and right limits are not equal at \( \pi/2 \), \( \lim_{x \to \pi/2} f(x) \) does not exist.

Evaluating these limits helps reveal where a function remains continuous or where its behavior shifts dramatically.

Floor Function

The floor function, represented often by brackets as in \( [ \cos x ] \), is a mathematical function stepping down non-integer values to their nearest lower integer. For example, if \( \cos x = 0.9 \), \( [ \cos x ] = 0 \).
This concept is critical here because \( f(x) = [ \cos x ] \) adjusts every point on the cosine wave to match its closest lower integer. It results in a 'stair-step' pattern - a hallmark characteristic of floor functions.
This transformation is pivotal because it converts continuous behavior into discrete jumps, which leads to interesting behaviors like discontinuity, which is common in step functions due to instantaneous value changes.

Discontinuity in Functions

Discontinuity occurs when a function shows an abrupt change at certain points, disrupting a smooth graph. For \( f(x) = [ \cos x ] \), discontinuities arise from how the floor function operates.
In this exercise, discontinuities occur at \( x = \pm \pi/2 \). As \( \cos x \) passes through integer values (here at zero), \( f(x) \) makes an instantaneous jump. The limits from either side of these points are unequal, causing a break — a characteristic discontinuity.
It’s these jumps where the function does not settle on a singular value. Recognizing such areas (odd multiples of \( \pi/2 \)) is crucial for understanding how and where \( \lim_{x\to a} f(x) \) exists or does not. At all non-discontinuous points, the step function behaves predictably, and the limit exists.