Question

Find the indicated limit or state that it does not exist. \(\lim _{(x, y) \rightarrow(2, \pi)}\left[x \cos ^{2}(x y)-\sin (x y / 3)\right]\)

Short answer

The limit is \(\frac{4 - \sqrt{3}}{2}\).

Step-by-step solution

Verify continuity of functions involved

The expression inside the limit, \([x \cos^2(xy) - \sin(xy/3)],\) consists of standard trigonometric functions. If these functions are continuous at a point, the limit can be evaluated by direct substitution. Cosine and sine functions are continuous everywhere, and multiplication and subtraction are continuous in their domains as well.

Substitute the given point into the expression

Since the trigonometric functions and algebraic operations involved are continuous, we can directly substitute \(x = 2\) and \(y = \pi\) into the expression. Calculate each term separately:\[ x \cos^2(xy) = 2 \cos^2(2 \cdot \pi) = 2 \cos^2(2\pi) = 2 \times 1^2 = 2 \].

Calculate the second term

Now calculate \(-\sin(xy/3)\):\(sin(2 \cdot \pi/3) = \sin(\frac{2\pi}{3}) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\)Thus, the second term becomes \(-\frac{\sqrt{3}}{2}\).

Compute the limit

Combine the results of the previous steps:\[ 2 - \frac{\sqrt{3}}{2} = \frac{4}{2} - \frac{\sqrt{3}}{2} = \frac{4 - \sqrt{3}}{2}\]. Thus, the limit is \frac{4 - \sqrt{3}}{2}\.

Key concepts

Continuity

In calculus, when we approach limits, one essential concept to understand is continuity. A function's continuity at a point means there are no interruptions, jumps, or gaps at that point. This is crucial because continuity allows us to determine limits easily by using substitution.
In the given limit problem, the expression involves trigonometric functions, cosine and sine, which are known for being continuous everywhere on the real line. This means they do not have any breaks or undefined points. As a result, when a function is continuous, it simplifies the process of finding a limit to the point where we can directly substitute the values of variables into the function.
Ensure that you check the continuity of all components of your expression to verify the existence of a limit. If they are continuous, then you can proceed with direct substitution without worrying about undefined behaviors.

Trigonometric Functions

Trigonometric functions like sine and cosine are common in calculus, especially when dealing with periodic phenomena or angles. Each of these functions can create oscillating outputs that rely on the input's interaction with pi.
In our problem,- \( \cos^2(xy) \) means that we apply the cosine function to the product of \( x \) and \( y \), then square it.
- Similarly, \( \sin(xy/3) \) takes the sine of \( xy/3 \), which involves both multiplication and division before using the sine function.Understanding how these functions transform values helps in computing limits. Cosine and sine values are predictable, ranging between -1 and 1. Thus, taking cosines and sines of numerous angles results in cycles and behavior we can predict or manipulate, even as inputs become complex or involve constants like pi.
To find the limits of expressions involving these functions, substitute the necessary points and use identities to simplify when needed.

Direct Substitution

Direct substitution is a straightforward method to solve limits in calculus. It involves replacing variables in a function with given values directly, which is possible only if the function is continuous at that point.
Since the functions involved in the problem \( x \cos^2(xy) - \sin(xy/3) \) are continuous when they approach specific values, direct substitution works effectively. By putting \( x = 2 \) and \( y = \pi \) into the function, you calculate stepwise:- Evaluate each part separately, for example, \( 2 \cos^2(2\pi) = 2 \times 1 = 2 \) because \( \cos(2\pi) = 1 \).
- Similarly, calculate \( -\sin(2\pi/3) \), utilizing trigonometric identities, to get \(-\frac{\sqrt{3}}{2} \).
After substitution, combine these separate evaluations to obtain the final limit.Using direct substitution simplifies limits considerably, avoiding more intricate calculus techniques like L'Hopital's rule or epsilon-delta proofs, provided continuity is guaranteed.