Problem 70
Question
Given \(f(x)=\cos (\tan x),\) find $$\lim _{x \rightarrow \frac{\pi}{2}} f(x)$$
Step-by-Step Solution
Verified Answer
The limit does not exist because \(\cos(\tan x)\) oscillates between -1 and 1.
1Step 1: Understanding the Function
We are given the function \( f(x) = \cos(\tan x) \). We need to find the limit as \( x \) approaches \( \frac{\pi}{2} \). The function comprises two parts: the tangent function and the cosine function. As we approach \( x = \frac{\pi}{2} \), we must understand the behavior of \( \tan x \) and \( \cos \) for those values of \( x \).
2Step 2: Evaluate the Behavior of \(\tan x\)
The tangent function, \( \tan x \), has a vertical asymptote at \( x = \frac{\pi}{2} \). As \( x \) approaches \( \frac{\pi}{2} \) from the left, \( \tan x \) increases towards positive infinity. From the right, \( \tan x \) decreases towards negative infinity.
3Step 3: Consider the Properties of \(\cos x\)
The cosine function, \( \cos(\theta) \), is defined and oscillates between -1 and 1 for all real \( \theta \). No matter how large or how negative \( \theta \) becomes, \( \cos(\theta) \) will always be within the interval [-1, 1].
4Step 4: Approach \(\lim_{x \to \frac{\pi}{2}} \cos(\tan x)\)
Given the nature of \( \tan x \) as \( x \to \frac{\pi}{2} \), \( \tan x \) tends to \( \pm \infty \). However, \( \cos(\theta) \) does not approach a limit as \( \theta \) goes to infinity or negative infinity, but rather oscillates between -1 and 1. This means that \( \cos(\tan x) \) does not settle at any specific value as \( x \to \frac{\pi}{2} \).
5Step 5: Conclusion on the Limit
Since \( \cos(\tan x) \) does not converge to any specific value but continues to oscillate between -1 and 1 as \( x \to \frac{\pi}{2} \), the limit \( \lim_{x \to \frac{\pi}{2}} f(x) \) does not exist.
Key Concepts
Trigonometric FunctionsAsymptotesFunction BehaviorInfinite Oscillation
Trigonometric Functions
Understanding trigonometric functions is key to solving limits involving them. In this exercise, we work with two trigonometric functions: tangent and cosine.
- Tangent Function (\(\tan x\)): This function represents the ratio of the sine to the cosine of an angle. As you approach special points like \(\frac{\pi}{2}\), the function's behavior becomes unique with significant changes.
- Cosine Function (\(\cos x\)): This function measures the horizontal length of a point on the unit circle moving through an angle. It's periodic, with values that repeat every \(2\pi\). The cosine value is always between -1 and 1, showing a regular, predictable pattern, unlike tangent.
Asymptotes
An asymptote refers to a line that a graph approaches but never actually reaches as the input values become extreme.
Tangent functions have vertical asymptotes where the function grows towards positive or negative infinity. Specifically, \(\tan x\) has a vertical asymptote at each odd multiple of \(\frac{\pi}{2}\), such as at \(x = \frac{\pi}{2}\). This means as \(x\) approaches \(\frac{\pi}{2}\) from either side, \(\tan x\) becomes unbounded:
Tangent functions have vertical asymptotes where the function grows towards positive or negative infinity. Specifically, \(\tan x\) has a vertical asymptote at each odd multiple of \(\frac{\pi}{2}\), such as at \(x = \frac{\pi}{2}\). This means as \(x\) approaches \(\frac{\pi}{2}\) from either side, \(\tan x\) becomes unbounded:
- Approaches positive infinity from the left.
- Approaches negative infinity from the right.
Function Behavior
The behavior of a function helps determine its limits as the input approaches certain values. When analyzing \(f(x) = \cos(\tan x)\), understanding the interplay between tangent and cosine is essential.
- Tangent's Extreme Behavior: As you approach \(x = \frac{\pi}{2}\), \(\tan x\) exhibits very rapid changes, going from negative infinity to positive infinity.
- Cosine's Oscillation: Although \(\tan x\) assumes extreme values, \(\cos(\tan x)\) simply oscillates between -1 and 1 without settling. It doesn't follow a straightforward growth or decay pattern.
Infinite Oscillation
Infinite oscillation indicates a function fluctuates continuously without approaching a finite limit.
For \(f(x) = \cos(\tan x)\), as \(x\) approaches \(\frac{\pi}{2}\), \(\tan x\) shoots to infinite positive and negative values. Consequently, \(\cos(\tan x)\) endlessly cycles through its range of -1 to 1.
For \(f(x) = \cos(\tan x)\), as \(x\) approaches \(\frac{\pi}{2}\), \(\tan x\) shoots to infinite positive and negative values. Consequently, \(\cos(\tan x)\) endlessly cycles through its range of -1 to 1.
- This persistent fluctuation between -1 and 1 creates an indefinite pattern.
- Without stabilization towards any single value, the limit is non-existent.
Other exercises in this chapter
Problem 70
Evaluate the limits using limit properties. If a limit does not exist, state why. $$\lim _{x \rightarrow 1} \frac{x^{2}-4 x+3}{x^{2}-2 x+1}$$
View solution Problem 70
For Exercises 65 through 70 , evaluate each limit. $$\lim _{x \rightarrow-\infty} \frac{\sqrt[3]{216 x^{3}+36 x^{2}-6 x+1}}{2 x}$$
View solution Problem 71
Evaluate the limits using limit properties. If a limit does not exist, state why. $$\lim _{x \rightarrow-7} \frac{\sqrt{x^{2}+14 x+49}}{x^{2}+8 x+7}$$
View solution Problem 71
Use the rational zeroes theorem to write the polynomial in completely factored form: \(3 x^{4}-19 x^{3}+15 x^{2}+27 x-10\)
View solution