Problem 1
Question
In Exercises \(1-8,\) use graphs and tables to find (a) \(\lim _{x \rightarrow \infty} f(x)\) and (b) \(\lim _{x \rightarrow-\infty} f(x)\) (c) Identify all horizontal asymptotes. $$f(x)=\cos \left(\frac{1}{x}\right)$$
Step-by-Step Solution
Verified Answer
The limits as \(x\) approaches infinity and negative infinity do not exist as the function oscillates between -1 and 1. The horizontal asymptotes for the function are \(y = 1\) and \(y = -1\).
1Step 1: Understanding the behaviour of the function
The function \(f(x)=\cos \left(\frac{1}{x}\right)\) becomes more oscillatory as \(x\) approaches 0. Cosine function oscillates between -1 and 1, hence for large \(x\), the value of \(f(x)\) will oscillate between -1 and 1.
2Step 2: Finding the limit as \(x\) approaches infinity
As \(x\) becomes larger, \(\frac{1}{x}\) becomes closer to 0. Hence, the function tends to oscillate faster and faster. Thus, the limit as \(x\) approaches infinity is undefined because the function does not approach a particular value but oscillates between -1 and 1.
3Step 3: Finding the limit as \(x\) approaches negative infinity
Similarly, as \(x\) becomes infinitely negative, our function also oscillates between -1 and 1 without settling down to a particular number. So the limit as \(x\) approaches negative infinity is also undefined.
4Step 4: Identify horizontal asymptotes
A horizontal asymptote is a horizontal line that the curve approaches as \(x\) tends to infinity or negative infinity. Since function \(f(x)\) oscillates between -1 and 1, the lines \(y = 1\) and \(y = -1\) are the horizontal asymptotes.
Key Concepts
Horizontal AsymptotesInfinite LimitsOscillating Functions
Horizontal Asymptotes
Horizontal asymptotes are flat lines that a function approaches as the input variable, usually named \(x\), heads off to infinity or negative infinity. They're great for understanding the long-term behavior of a function. In our exercise, the function \(f(x) = \cos\left(\frac{1}{x}\right)\) has horizontal asymptotes. Even though \(f(x)\) doesn't settle at a single value as \(x\) heads towards infinity, it does hover between two boundaries. This suggests the presence of horizontal asymptotes.
- For \(f(x) = \cos\left(\frac{1}{x}\right)\), the cosine part ranges between -1 and 1.
- As \(x\) becomes very large or very small, \(\frac{1}{x}\) gets closer to zero, making the function \(\cos\left(\frac{1}{x}\right)\) oscillate between -1 and 1.
- Thus, the horizontal asymptotes are \(y = -1\) and \(y = 1\).
Infinite Limits
Infinite limits involve the behavior of a function as \(x\) approaches infinity or negative infinity. It's like asking, "What happens to the function as we go really far along the \(x\)-axis?" In the context of our function \(f(x) = \cos\left(\frac{1}{x}\right)\), things get interesting.
- As \(x\) tends to infinity, \(\frac{1}{x}\) shrinks towards zero. But, since cosine oscillates, \(f(x)\) continually shifts between -1 and 1.
- Unlike many functions which may approach a specific number, \(f(x)\) doesn't settle at a constant value.
- This means that the limits – \(\lim_{x \to \infty} f(x)\) and \(\lim_{x \to -\infty} f(x)\) – are both undefined.
Oscillating Functions
Oscillating functions are those that move back and forth between certain bounds without settling on a particular value. The function \(f(x)=\cos\left(\frac{1}{x}\right)\) is a perfect example. It demonstrates fascinating traits as \(x\) approaches both positive and negative extremes
- The term \(\frac{1}{x}\) means that as \(x\) grows, its reciprocal shrinks toward zero, causing \(\cos\left(\frac{1}{x}\right)\) to cycle rapidly between -1 and 1.
- This cyclic nature implies no steady limit at infinity or negative infinity, yet hints at a structured pattern.
- Despite its perpetual oscillation, the function stays confined within fixed bounds: -1 and 1.
Other exercises in this chapter
Problem 1
In Exercises \(1 - 4 ,\) an object dropped from rest from the top of a tall building falls \(y = 16 t ^ { 2 }\) feet in the first \(t\) seconds. Find the averag
View solution Problem 1
In Exercises \(1-10,\) find the points of continuity and the points of discontinuity of the function. Identify each type of discontinuity. $$y=\frac{1}{(x+2)^{2
View solution Problem 2
In Exercises \(1-6,\) find the average rate of change of the function over each interval. \(f(x)=\sqrt{4 x+1}\) (a) $$[0,2] \quad$$ (b) $$[10,12]$$
View solution Problem 2
In Exercises \(1 - 4 ,\) an object dropped from rest from the top of a tall building falls \(y = 16 t ^ { 2 }\) feet in the first \(t\) seconds. Find the averag
View solution