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Find the limit. (Hint: Let \(x=1 / t\) and find the limit as \(t \rightarrow 0^{+}\).) $$ \lim _{x \rightarrow \infty} x \sin \frac{1}{x} $$

Short Answer

Expert verified
The limit of the given function \(x \sin \frac{1}{x}\) as \(x \rightarrow \infty\) is 1.

Step by step solution

01

- Introduce a new variable

According to the hint, let’s make a substitution: Let \(x=1/t\). Then, as \(x \rightarrow \infty\), it follows that \(t \rightarrow 0^{+}\). So, our limit will be modified as follows: \(\lim _{x \rightarrow \infty} x \sin \frac{1}{x}\) become \(\lim _{t \rightarrow 0^{+}} \frac{\sin t}{t}\).
02

- Applying L'Hopital’s rule

We can see that as \(t \rightarrow 0^{+}\), both the numerator and the denominator are going to zero. In this classic 0/0 indeterminate form, we apply L'Hopital's rule, which states that the limit of the ratio of two functions as they go to zero or infinity is the same as the limit of the ratio of their derivatives. Hence, \(\lim _{t \rightarrow 0^{+}} \frac{\sin t}{t}\) is equal to \(\lim _{t \rightarrow 0^{+}} \frac{\cos t}{1}\), after differentiating the sine and \(t\) with respect to \(t\).
03

- Evaluate the limit

Now, it's straight forward to substitute the limit in \(\lim _{t \rightarrow 0^{+}} \frac{\cos t}{1}\), which becomes \(\cos 0 = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hopital's Rule
Understanding the intricacies of L'Hopital's Rule is crucial while studying calculus, especially when encountering indeterminate forms. This rule provides a method to evaluate limits of ratios where direct substitution is not possible due to an indeterminate form such as 0/0. Applied correctly, L'Hopital's Rule states that if the limit of the functions in the numerator and denominator are zero or infinity, we can find the limit of the original function by taking the derivative of both the numerator and the denominator separately, and then finding the limit of the new ratio.

For instance, in the given problem, the limit of \( \frac{\text{sin } x}{x} \) as \( x \) approaches infinity translates into a 0/0 scenario when \( t \rightarrow 0^{+} \) after the substitution \( x=1/t \) is made. Therefore, applying L'Hopital's Rule by differentiating the numerator and denominator leads us to a new limit: \( \frac{\text{cos } t}{1} \) where \( t \) approaches 0. L'Hopital's Rule simplifies the evaluation process, transforming a seemingly complicated limit into an easily calculable one.

However, it's also important to note that L'Hopital's Rule can only be applied when certain conditions are met, and sometimes it needs to be applied more than once, especially if the result after the first differentiation is still an indeterminate form.
Indeterminate Forms
Indeterminate forms are mathematical expressions whose behavior is not immediately obvious; the most common forms being 0/0 and \( \frac{\text{∞}}{\text{∞}} \). In calculus, these emerge often when evaluating limits. It's vital to remember that an indeterminate form does not indicate the limit does not exist, rather, it requires further manipulation to find a definitive answer.

In the given exercise, we encounter a 0/0 form when applying the substitution method and finding the limit as \( t \) approaches 0. Traditionally, calculus students are taught to handle such scenarios with strategies like factoring, simplifying, or using trigonometric identities. However, when these strategies don't work, L'Hopital's Rule becomes an effective solution.

Indeterminate forms can also take other versions, such as \( 0 \times \text{∞} \), \( \text{∞} - \text{∞} \), \( 0^{0} \), \( \text{∞}^{0} \), and \( 1^{\text{∞}} \), each presenting unique challenges. Recognizing when an expression is in an indeterminate form is essential as it indicates an opportunity for further exploration to reveal the limit's true value.
Trigonometric Limits
Trigonometric limits are a subset of limits in calculus that specifically involve trigonometric functions. They often feature in problems dealing with periodic functions and can exhibit indeterminate forms like the one examined in our textbook exercise, \( \text{sin} \( t \) / t \), as \( t \) approaches zero.

When it comes to trigonometric limits, understanding certain foundational limits is a game-changer. For example, the limit of \( \text{sin} \( x \) / x \) as \( x \) approaches 0 is well-known to be 1. This is a critical tool and serves as a basis for solving many other trigonometric limits. Furthermore, knowing the behavior of trigonometric functions around key angles can greatly simplify the evaluation process, without necessarily invoking a heavy-hitting tool like L'Hopital's Rule.

In our problem, the substituting method led us to the limit of \( sin \( t \) / t \), a trigonometric limit. Knowing its behavior at the angle 0 and applying L'Hopital's Rule confirmed the limit as 1. Such knowledge of trigonometric limits is indispensable not only in pure mathematics but also in fields as diverse as physics and engineering.
Substitution Method
The substitution method in calculus is a technique used to simplify complex limit problems. By substituting one variable for another, often through a well-thought-out expression, we can transform the limit into one that is more straightforward to handle. This method is particularly useful when we encounter functions that are difficult to evaluate using direct substitution.

For the given exercise, the substitution \( x=1/t \) as \( x \) approaches infinity, which then leads to \( t \) approaching 0 from the positive side, is a clever tactic. It switches the function from involving an infinite variable \( x \) to a variable \( t \) approaching zero, enabling the use of well-known limits and rules, like L'Hopital's Rule.

Successful application of the substitution method requires familiarity with limits and keen insight into the function's behavior. The effectiveness of this technique is seen in how it simplifies finding the limit of \( x \text{sin} \( 1/x \) \) as \( x \) approaches infinity—what seems like a daunting task at first becomes an elegant solution through substitution, showcasing the power of this mathematical tool.

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