Problem 107 - Calculus Volume 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, we have that \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Hence, \( \sqrt{x^2 + y^2} = r \). The given function becomes \( \frac{\sin(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} = \frac{\sin(r)}{r} \).

Step 2: Consider the Limit in Terms of r

We want to find \( \lim _{(x, y) \rightarrow (0,0)} \frac{\sin(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} \), which is equivalent to \( \lim _{r \rightarrow 0} \frac{\sin(r)}{r} \) since the direction \( \theta \) does not affect the limit in polar coordinates.

Step 3: Apply the Sine Limit Theorem

A well-known limit in calculus is \( \lim_{r \to 0} \frac{\sin(r)}{r} = 1 \). This is because as \( r \) approaches 0, the sine of \( r \) over \( r \) approaches 1.

Step 4: State the Final Conclusion

Thus, the limit is given by \( \lim _{(x, y) \rightarrow (0,0)} \frac{\sin(\sqrt{x^2 + y^2})}{\sqrt{x^2 + y^2}} = 1 \).

Key Concepts

Understanding L'Hôpital's Rule

L'Hôpital's rule is a powerful tool in calculus for evaluating limits that result in indeterminate forms. An indeterminate form is an expression that doesn't have a clear or obvious value, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule allows us to differentiate the numerator and the denominator separately and then re-evaluate the limit.

Here's how it works:

• Check the limit initially to see if it gives an indeterminate form.
• If it does, differentiate the top and bottom separately.
• Recalculate the limit with these new derivatives.
• If the result is still indeterminate, you can apply the rule again.

L'Hôpital's rule is especially useful when dealing with trigonometric, exponential, or logarithmic functions, which can be complex to decipher directly. However, ensure the original limit is indeed indeterminate before applying the rule, as it's not needed otherwise.

Exploring the Sine Limit Theorem

The sine limit theorem is crucial for understanding trigonometric limits. This theorem states that \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\). It's an essential part of calculus when working with trigonometric functions approaching zero.

The intuition behind this theorem is straightforward:

• As \(x\) becomes very small, the angle \(x\) in radians and \(\sin(x)\) approach the same value.
• This causes the ratio of \(\frac{\sin(x)}{x}\) to converge to 1.

This limit is often proven using geometric arguments, where one can visualize the unit circle. A key learning point is recognizing the difference between angles measured in degrees versus radians, with this theorem applying strictly in radian measure.

The Role of Calculus in Problem Solving

Calculus is a branch of mathematics that deals primarily with change and motion and helps us solve complex problems involving rates of change and areas under curves. Its two main branches are differentiation and integration.

Why is calculus important?

• Calculus allows for the calculation of limits, essential for understanding continuous change.
• It provides tools, like L'Hôpital's rule and the sine limit theorem, for evaluating difficult limits.
• Calculus is used in many fields such as physics, engineering, economics, and any areas that involve dynamic systems.

In any calculus problem, identifying which rules or theorems apply is crucial. For example, in the exercise here, polar coordinates simplify the problem, and applying the sine limit theorem ensures the accurate computation of the limit.