Problem 35 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The limit \( \lim_{x \rightarrow \infty} \sqrt{x} \sin \frac{1}{x} = 0 \).

Step 1: Identify the limit form

Examine the expression \( \sqrt{x} \sin \frac{1}{x} \) as \( x \rightarrow \infty \). The integral parts are \( \sqrt{x} \) growing unbounded and \( \sin \frac{1}{x} \rightarrow \sin(0) = 0 \). The form is therefore \( \infty \cdot 0 \) which is indeterminate.

Step 2: Simplify using substitution

To resolve the indeterminate form, set a substitution: let \( u = \frac{1}{x} \), hence \( x = \frac{1}{u} \). When \( x \rightarrow \infty \), then \( u \rightarrow 0^+ \). The expression becomes \( \frac{1}{\sqrt{u}} \sin(u) \). Now, find the limit \( \lim_{u \rightarrow 0^+} \frac{\sin(u)}{\sqrt{u}} \).

Step 3: Application of l'Hôpital's rule

The expression \( \frac{\sin(u)}{\sqrt{u}} \) as \( u \rightarrow 0^+ \) assumes a \( \frac{0}{0} \) form. Utilize l'Hôpital's rule by differentiating the numerator and denominator. The derivative of \( \sin(u) \) is \( \cos(u) \) and that of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \). The limit transforms to: \( \lim_{u \rightarrow 0^+} \frac{\cos(u)}{\frac{1}{2\sqrt{u}}} = \lim_{u \rightarrow 0^+} 2\sqrt{u}\cos(u) \).

Step 4: Evaluate the transformed limit

Evaluate \( \lim_{u \rightarrow 0^+} 2\sqrt{u}\cos(u) \). As \( u \rightarrow 0^+ \), \( \sqrt{u} \rightarrow 0 \) and \( \cos(u) \rightarrow 1 \). Therefore, the overall expression \( 2\sqrt{u}\cos(u) \rightarrow 2\cdot0\cdot1 = 0 \).

Step 5: Conclude the limit

Having evaluated through substitution and applying l'Hôpital's rule, we conclude the limit is \( 0 \). Hence, the original expression \( \lim_{x \rightarrow \infty} \sqrt{x} \sin \frac{1}{x} \) simplifies to \( 0 \).

Key Concepts

Indeterminate Forms

When dealing with limits, you might encounter forms where the outcome is not immediately clear. These are known as "Indeterminate Forms." An indeterminate form suggests that the limit doesn't have an obvious numerical value without further analysis. When we encounter such forms, like \( \infty \cdot 0 \) in our exercise, it's crucial to proceed with caution and employ methods such as substitution or l'Hôpital's Rule to understand the true value of the limit.

Indeterminate forms can appear in various expressions:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( \infty - \infty \)
• \( 0 \cdot \infty \)
• \( 0^0 \)
• \( \infty^0 \)
• \( 1^\infty \)

Recognizing that an expression is indeterminate allows us to apply mathematical tools to resolve the ambiguity and evaluate the limit effectively. It is a fundamental concept in calculus, helping students understand that not all mathematical expressions have straightforward solutions.

l'Hôpital's Rule

l'Hôpital's Rule is a powerful tool in calculus used to resolve indeterminate forms, particularly \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It provides a method to find the limit of these expressions by differentiating the numerator and the denominator separately.

Here's how l'Hôpital's Rule works:

• Identify that the limit is an indeterminate of form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiating the numerator and the denominator until the expression no longer results in an indeterminate form.
• Evaluate the new limit, which should now be determinable.

In our exercise, after substitution, the expression \( \frac{\sin(u)}{\sqrt{u}} \) resulted in a \( \frac{0}{0} \) indeterminate form. By differentiating \( \sin(u) \) to \( \cos(u) \) and \( \sqrt{u} \) to \( \frac{1}{2\sqrt{u}} \), we transformed the problem into a more straightforward limit, easily evaluated as \( 0 \). This rule simplifies the process of finding limits in otherwise complex expressions.

Substitution Method

The Substitution Method is a strategic technique used to simplify complex limits and resolve indeterminate forms by changing variables. This approach can sometimes convert a difficult limit into one that's simpler to evaluate.

Here's a simple breakdown of the Substitution Method:

• Identify parts of the expression that, when re-expressed in terms of a new variable, simplify the evaluation.
• Choose a suitable substitution that simplifies the expression into a determinable form.
• Apply the limit process to the new variable.
• Substitute back if necessary to relate to the original terms.

In our example, setting \( u = \frac{1}{x} \) and thus transforming \( x \) to \( \frac{1}{u} \) allowed us to rewrite the limit expression in terms of \( u \). This changed the problem from \( \infty \cdot 0 \) to a \( \frac{0}{0} \) scenario, which could then be resolved using l'Hôpital's Rule, ultimately simplifying the solution process. It's a versatile technique that complements other calculus methods to tackle challenging limit problems efficiently.