Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right) $$

Short answer

The limit is 1.

Step-by-step solution

Analyze the Limit Form

First, substitute the variable tending towards infinity in the given limit expression. Observe that as \( x \) approaches infinity, the expression \( \frac{1}{x} \) approaches 0. Thus, \( x \sin\left(\frac{1}{x}\right) \) can be rewritten as a product of a very large number \( x \) and a number \( \sin\left(\frac{1}{x}\right) \) which approaches 0. This forms an indeterminate form \( \infty \cdot 0 \). This suggests using a different method or performing algebraic manipulation to simplify the limit.

Rewrite the Expression

To simplify the evaluation, express the limit in a different form. Rewrite \( x \sin\left(\frac{1}{x}\right) \) as \( \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}} \). This transforms the limit into a \( \frac{0}{0} \) form and makes it suitable for L'Hôpital's rule.

Apply L'Hôpital's Rule

According to L'Hôpital's rule, if the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) is indeterminate in the form \( \frac{0}{0} \), then it is equal to \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \). Therefore, we differentiate both the numerator and denominator:- Numerator: Differentiate \( \sin\left(\frac{1}{x}\right) \) with respect to \( x \). Using the chain rule, the derivative is \( \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = -\frac{\cos\left(\frac{1}{x}\right)}{x^2} \).- Denominator: Differentiate \( \frac{1}{x} \) with respect to \( x \), which is \(-\frac{1}{x^2} \).

Simplify and Evaluate the Limit

Apply the derivatives to find the new limit: \[\lim_{x \to \infty} \frac{-\frac{\cos\left(\frac{1}{x}\right)}{x^2}}{-\frac{1}{x^2}} = \lim_{x \to \infty} \cos\left(\frac{1}{x}\right).\]As \( x \to \infty \), \( \frac{1}{x} \to 0 \), hence \( \cos\left(\frac{1}{x}\right) \to \cos(0) = 1 \). Thus, the limit evaluates to 1.

Key concepts

Limit Evaluation

When faced with a limit problem, we're trying to find what a function approaches as the input reaches a certain value. This is a fundamental concept in calculus. In our exercise, we're interested in what happens to the expression \(x \sin\left(\frac{1}{x}\right)\) as \(x\) becomes infinitely large. Let's break it down:

• Start by substituting the value the variable is approaching into the expression. This helps to understand the behavior of the function.
• If you encounter a problematic form, like our \(\infty \cdot 0\), you'll need to manipulate the expression to find a solution.
• In this case, rewriting the original limit as \(\lim_{x \to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\) transforms it into a new, more manageable format.

Evaluating limits often requires creativity in changing the form of the function to make it simpler to analyze.

Indeterminate Form

An indeterminate form arises when substitution in a limit problem does not directly give an answer, but rather a problem that needs further analysis. Examples include forms like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and \(\infty \cdot 0\). In our exercise, when \(x\) is very large, \(\sin\left(\frac{1}{x}\right)\) becomes very small, creating an \(\infty \cdot 0\) form.Dealing with such a form means:

• Determining a new form of the expression that allows evaluating the limit. For \(x \sin\left(\frac{1}{x}\right)\), we rewrote it to form \(\frac{0}{0}\).
• Using a tool like L'Hôpital's rule, which provides a method to handle these indeterminate cases effectively by differentiating the numerator and the denominator.

Such techniques make it possible to resolve the undetermined nature and find a finite limit.

Trigonometric Limits

Trigonometric limits involve functions like sine and cosine, which have unique properties that can be useful in calculus. In our problem, \(\sin\left(\frac{1}{x}\right)\) is a trigonometric function that becomes crucial.Here are some tips for dealing with trigonometric limits:

• Understand the behavior of trigonometric functions at key points, like \( \sin(0) = 0\) and \( \cos(0) = 1\). These basic results are powerful in limit evaluations.
• Use known limits such as \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\). This can be highly beneficial when functions appear in transformed forms.
• These properties were key in our problem, where by rewriting the expression, we leverage the limit property of \(\cos\) as it approached 0.

Trigonometric limits often require recognizing these patterns and applying their unique properties to solve related problems.