Problem 107 - Calculus 3

Question

Use polar coordinates to find \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin \sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}},\) You can also find the limit using L'Hôpital's rule.

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Convert to Polar Coordinates

In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \), where \( r = \sqrt{x^2 + y^2} \). Rewriting the expression, we get \( \lim _{r \to 0} \frac{\sin r}{r} \), which is easier to evaluate.

Step 2: Apply the Known Limit Property

The limit \( \lim _{r \to 0} \frac{\sin r}{r} \) is a standard limit in calculus, and its value is known to be 1. This is a direct limit evaluation without requiring further manipulation.

Step 3: Conclude the Limit Evaluation

Since the expression simplifies directly to a known limit property in standard calculus, the limit of the original function as \((x, y) \to (0,0)\) is simply \(1\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a valuable mathematical tool used to evaluate limits involving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When limits present these forms, it can be challenging to find straightforward solutions. L'Hôpital's Rule states that if you have a function \( \lim_{x \to c} \frac{f(x)}{g(x)} \) that results in such an indeterminate form, you can differentiate \( f(x) \) and \( g(x) \) separately and then take the limit of their derivatives to find the original limit:

• If \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\infty}{\infty} \)
• Then, \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) will provide the sought limit, if it exists.

Using L'Hôpital's Rule simplifies many complex calculations, especially when functions are non-linear. Instead of directly linking to the given exercise, one might choose this rule as an alternate verification strategy. However, the problem provided is best approached using polar coordinates due to its multivariable nature, which circumvents the need for L'Hôpital's Rule.

Limit Evaluation

Limit evaluation is a process in calculus used to determine the value that a function approaches as the input approaches some point. Understanding limits is fundamental since they form the basis for defining derivatives and integrals. Different techniques are used for limit evaluation, including simplifying expressions, factoring, and using trigonometric identities.
For example:

• Direct substitution, where you simply plug the values into the function.
• Simplification, which includes factoring or reducing the expression.
• The Squeeze Theorem, which sandwiches the function between two others to find the limit.

In the context of the provided exercise, rewriting the function using polar coordinates makes it easier to directly evaluate the limit. Instead of handling each variable separately, using \( r \) gives a single variable approach, leading to a clearer solution.

Standard Limit Property

The standard limit property involves the recognition of frequently encountered limits in calculus, which are foundational for solving many problems quickly and insightfully. One of the most common is the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), which often appears in trigonometric and polar problems.

This standard property is derived from the unit circle in trigonometry, which relates the sine of a tiny angle to its measure in radians. As a key result:

• This property allows students to bypass more complex calculus methods.
• It is utilized without further proof due to its widespread understanding and proof existence in calculus courses.

In the exercise, converting to polar coordinates reduced the expression to this standard limit property. Knowing and applying these properties streamlines the process, offering a quick solution to otherwise challenging limits.