Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The limit is 0 as \(x\to 0\).

Step 1: Analyze the Expression

We need to find the limit \(\lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\tan x}\). Start by checking the behavior of \(\sin(1/x)\) at \(x=0\). As \(x\rightarrow 0, \sin(1/x)\) oscillates between -1 and 1. However, since \(x^2\) tends to 0 as \(x\to 0\), our task simplifies to analyzing the limit\[\lim_{x\to 0} \frac{x^2 \sin(1/x)}{\tan x}\].

Step 2: Approximations Near Zero

For small \(x\), the tangent function \(\tan x\) can be approximated by \(x\) itself, i.e., \(\tan x \approx x\). This gives us:\[\frac{x^2 \sin(1/x)}{\tan x} \approx \frac{x^2 \sin(1/x)}{x} = x \sin(1/x).\]

Step 3: Evaluate the Limit of the Simpler Expression

Now, consider the limit of the expression \(x \sin(1/x)\) as \(x\to 0\). Since \(\sin(1/x)\) is bounded between -1 and 1, the product \(x \sin(1/x)\) will be bounded by \(-x\) and \(x\). As \(x\to 0\), both \(-x\) and \(x\) tend to 0.

Step 4: Apply the Squeeze Theorem

Since \(-x \leq x \sin(1/x) \leq x\), and \(-x\to 0\) and \(x\to 0\) as \(x\to 0\), we can use the squeeze theorem to conclude that:\[\lim_{x \to 0} x \sin(1/x) = 0.\]

Step 5: Conclusion of the Original Limit

Hence, the original limit \(\lim _{x \rightarrow 0} \frac{x^{2} \sin (1 / x)}{\tan x}\) simplifies to \(0\) using the squeeze theorem on the simpler bound expression.

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus used for finding limits, especially when direct substitution doesn't work. It leverages the idea of bounding a problematic function between two simpler functions whose limits are known.

Imagine you have three functions, say \( f(x) \), \( g(x) \), and \( h(x) \), and we know that \( f(x) \leq g(x) \leq h(x) \) for all \( x \) in some interval around a point (excluding possibly at the point itself). If both \( f(x) \) and \( h(x) \) approach the same limit \( L \) as \( x \) approaches a certain value, then by the Squeeze Theorem, \( g(x) \) must also approach \( L \).

Let's see why this makes sense. If both upper and lower bounds squish the middle function to converge to the same point, the middle one has nowhere else to go. It's like squeezing a ball between your hands; it can't go anywhere else besides where your hands lead it. This is exactly the strategy applied in finding \( \lim_{x \to 0} \frac{x^2 \sin(1/x)}{\tan x} \). By bounding \( x \sin(1/x) \) between \( -x \) and \( x \), and knowing that both \( -x \rightarrow 0 \) and \( x \rightarrow 0 \) as \( x \rightarrow 0 \), the Squeeze Theorem tells us that \( x \sin(1/x) \) must also go to zero.

Trigonometric Functions

Trigonometric functions, such as sine and tangent, appear frequently in limits problems due to their periodic and oscillatory nature. This can make them challenging to evaluate directly. Instead, understanding their behavior can provide meaningful insights.

Take \( \sin(x) \), for example, which oscillates between -1 and 1. This quality makes it bound within these limits regardless of its argument, like \( \sin(1/x) \). On the other hand, \( \tan(x) \) has the property of being approximately equal to \( x \) when \( x \) is near zero. This approximation is crucial because it simplifies expressions involving \( \tan(x) \) when approaching limits.

In our original problem, using \( \tan x \approx x \) simplifies the expression \( \frac{x^2 \sin(1/x)}{\tan x} \) to \( x \sin(1/x) \), much more manageable when finding the limit. Understanding these approximations and behavior helps solve complex calculus problems, making trigonometric functions less daunting.

Oscillation and Limits

Oscillation is when a function fluctuates between values, often indefinitely. This characteristic can complicate limit problems. Yet, understanding how oscillation interacts with other functions can clarify these issues.

Consider \( \sin(1/x) \), which oscillates between -1 and 1 infinitely as \( x \) approaches zero. While \( \sin(1/x) \) never settles to a single value, when multiplied by \( x^2 \), the overall expression \( x^2 \sin(1/x) \) behaves differently due to the squaring influence.

The smaller \( x \) gets, the closer \( x^2 \) is to zero, regardless of the oscillation of \( \sin(1/x) \). Hence, the multiplication acts like a dampener, bringing the product towards zero, thereby allowing the application of the Squeeze Theorem effectively. Grasping oscillation's role in limit convergence can help you unwrap more scrambled limit cases.