Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { 6. Find } \lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\tan x} \text {. } $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Analyze the Limit Expression

First, observe the expression \( \frac{x^2 \sin(1/x)}{\tan x} \) as \( x \to 0 \). The numerator \( x^2 \sin(1/x) \) oscillates between \( -x^2 \) and \( x^2 \), and the denominator \( \tan x \approx x \) for \( x \) close to 0.

Step 2: Apply Limit Properties

Since \( \tan x \approx x \) when \( x \) approaches 0, we can approximate the expression by \( \lim_{x \to 0} \frac{x^2 \sin(1/x)}{x} \). This simplifies to \( \lim_{x \to 0} x \sin (1/x) \).

Step 3: Use Squeeze Theorem

We know \( -1 \leq \sin(1/x) \leq 1 \). Therefore, \( -x \leq x \sin(1/x) \leq x \). As \( x \to 0 \), both \( -x \) and \( x \) approach 0. By the squeeze theorem, \( x \sin(1/x) \to 0 \).

Step 4: Conclude the Result

Since \( x \sin(1/x) \to 0 \) as \( x \to 0 \), we conclude that the original limit \( \lim_{x \to 0} \frac{x^2 \sin(1/x)}{\tan x} = 0 \).

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool when dealing with limits that might appear challenging or complex. It helps us understand the behavior of a function by "squeezing" it between two simpler functions that have the same limit at a particular point. In this exercise, the key idea is that the trigonometric function \(\sin(1/x)\) oscillates between \(-1\) and \(1\), even as \(x\) gets very close to zero.

• We express this as: \(-1 \leq \sin(1/x) \leq 1\), which remains true for all values of \(x\) not equal to zero.
• If we multiply through by \(x\), we have: \(-x \leq x \sin(1/x) \leq x\).
• Finally, as \(x \to 0\), both \(-x\) and \(x\) approach \(0\).

This setup allows the use of the Squeeze Theorem. Since \(-x\) and \(x\) "squeeze" \(x \sin(1/x)\) to \(0\) when \(x\) approaches \(0\), we conclude that \(x \sin(1/x)\) goes to \(0\) as well.

Trigonometric Limits

Trigonometric limits often involve functions like \(\sin\), \(\cos\), and \(\tan\). Understanding their behavior as \(x\) approaches certain critical values (like \(0\)) is essential in calculus. In our case, the expression involves the term \(\sin(1/x)\), which behaves in a unique way.

• The function \(\sin(1/x)\) oscillates because the input \(1/x\) grows without bound as \(x\) gets close to \(0\).
• Despite the rapid oscillation, the range of \(\sin(1/x)\) is limited to \([-1, 1]\), which helps in simplifying complex limits with the Squeeze Theorem.
• Also, for small values of \(x\), \(\tan x\) can be approximated by \(x\) itself, simplifying limit problems.

These properties help us to conclude that the overall limit of the complex trigonometric expression can be easily estimated or proven.

Limit Properties

When working with limits, knowing certain properties can significantly simplify your calculations and understanding. Here are some key limit properties relevant in our exercise.

• As \(x\) approaches \(0\), the function \(\tan x\) is approximately equal to \(x\). This makes solving limits of expressions including \(\tan x\) easier.
• Any constant multiple of a function that approaches \(0\) will also approach \(0\). For example, if \(\lim_{x \to 0} x = 0\), then \(\lim_{x \to 0} x^2 = 0\).
• Finally, if a product \(f(x)g(x)\) approaches \(0\) from \(f(x) = 0\) and \(|g(x)|\) is bounded, then the product also approaches \(0\). In this case, \(x^2 \sin(1/x)\) simplifies to \(x \sin(1/x)\), using this property.

These properties are essential in navigating limits and simplifying complex expressions effectively.