Problem 36 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty} x^{2} \sin \frac{1}{x^{2}} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Limit Form

Notice that as \( x \to \infty \), \( \frac{1}{x^2} \to 0 \). This means the expression can be rewritten as \( x^2 \sin \left( \frac{1}{x^2} \right) \). We need to analyze the behavior of this expression as \( x \to \infty \).

Step 2: Approximate the Sine Function for Small Angles

As \( x \to \infty \), \( \frac{1}{x^2} \) becomes a very small number. For small angles \( \theta \), \( \sin \theta \approx \theta \). Hence, \( \sin \left( \frac{1}{x^2} \right) \approx \frac{1}{x^2} \).

Step 3: Substitute the Approximation

Substitute \( \sin \left( \frac{1}{x^2} \right) \) with \( \frac{1}{x^2} \) in the limit expression. The expression becomes \( x^2 \left( \frac{1}{x^2} \right) = 1 \).

Step 4: Evaluate the New Limit

Now, as \( x \to \infty \), since the expression becomes 1, the limit evaluates to \( 1 \). Thus, \( \lim _{x \rightarrow \infty} x^{2} \sin \frac{1}{x^{2}} = 1 \).

Key Concepts

l'Hôpital's rule

When tackling complex limit problems, l'Hôpital's Rule can be a valuable tool. It is particularly useful when dealing with indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). L'Hôpital's Rule states that for limits where direct substitution results in these forms, we can differentiate the numerator and denominator individually and then re-evaluate the limit using these derivatives. However, it is vital to ensure that the derivatives exist and that by substituting you again form the same type of limit. For the problem at hand, \( \lim_{x \to \infty} x^{2} \sin \left( \frac{1}{x^{2}} \right) \), there is no need to apply l'Hôpital's Rule. Instead, by recognizing the behavior of the trigonometric function for small angles, we bypass the need for l'Hôpital's Rule and directly address the limit. Using direct approximations can often simplify the computation, especially when dealing with the product of a polynomial and a trigonometric function.

infinite limits

Infinite limits describe the behavior of a function as the independent variable grows without bound, either positive or negative. In the context of this problem, as \( x \to \infty \), we examine how the function \( x^{2} \sin \left( \frac{1}{x^{2}} \right) \) behaves.

• Step 1 involves determining what happens to the expression inside the limit. We have that \( \frac{1}{x^2} \to 0 \) as \( x \to \infty \).
• In Step 2, understanding the nature of the sine function as \( x \) becomes very large is key. Small angle approximation, \( \sin \theta \approx \theta \), is used to simplify the calculation.
• This small angle approximation makes evaluating the limit straightforward by simplifying the trigonometric function to a polynomial evaluated at infinity, which often yields finite results.

The example shows mastering the concept of infinite limits often involves simplifying complex expressions by using approximations or recognizing essential properties of functions involved.

trigonometric functions

Trigonometric functions, like sine and cosine, are fundamental in calculus due to their periodic nature and relationships in various branches of mathematics. In limit problems, understanding these functions' behavior at key points, such as \( 0, \) and their intervals, is crucial.For very small angles \( \theta \), the sine function behaves linearly, which significantly simplifies computations. This linearization is typically expressed as the approximation \( \sin \theta \approx \theta \) when \( \theta \to 0 \). This approximation is particularly helpful in limit problems where substitution of longer, more intricate expressions by a simple variable is desired.In this exercise, \( \sin \left( \frac{1}{x^2} \right) \approx \frac{1}{x^2} \) makes the trigonometric component of the problem much easier to handle as \( x \to \infty \). This simplification ensures the correct evaluation of the limit and provides insight into approximating complicated trigonometric functions with simpler linear functions.