Problem 68 - Calculus 3

Question

For the following exercises, evaluate the limits at the indicated values of \(x\) and \(y\) . If the limit does not exist, state this and explain why the limit does not exist. $$ \lim _{(x, y) \rightarrow(0, \pi / 4)} \frac{\sec x+2}{3 x-\tan y} $$

Step-by-Step Solution

Answer

The limit is -3.

Step 1: Substitute Known Values into the Limit Expression

We know the limit is approaching \((x, y) \rightarrow (0, \pi/4)\). Substitute \(x = 0\) and \(y = \pi/4\) into the expression \(\frac{\sec x + 2}{3x - \tan y}\). This gives us \(\frac{\sec(0) + 2}{3(0) - \tan(\pi/4)}\).

Step 2: Calculate the Numerator

Evaluate \(\sec(0)\). Since \(\sec(0) = \frac{1}{\cos(0)} = 1\), the numerator becomes \(1 + 2 = 3\).

Step 3: Calculate the Denominator

Evaluate \(\tan(\pi/4)\). Since \(\tan(\pi/4) = 1\), the denominator becomes \(3(0) - 1 = -1\).

Step 4: Evaluate the Fraction

Combine the results from the numerator and the denominator to evaluate the fraction: \(\frac{3}{-1} = -3\).

Step 5: Conclusion About the Limit

Since the substitution gave a finite number without undefined operations, the limit exists. The value of the limit is \(-3\). This means the function approaches \(-3\) as \((x, y)\) approaches \((0, \pi/4)\).

Key Concepts

Limits of Functions

In multivariable calculus, limits of functions are foundational. Understanding how a function behaves as it approaches a specific point helps to analyze its properties. For functions of two variables, the limit describes how the function behaves as its inputs get closer to a particular pair of values.When dealing with limits in multiple dimensions, we examine the function value as \((x, y)\) approaches a point such as \((0, \frac{\pi}{4})\). This requires evaluating the function along every possible path to ensure consistency. If different paths yield different limits, the limit does not exist. However, if all paths result in the same value, the limit is valid and confirms that the function approaches a defined number. Understanding limits in functions is crucial for evaluating continuity and understanding the function's behavior around points of interest. By successfully evaluating the limit in our exercise, we determined that the function's limit existed, and the value was \(-3\) at the given point.

Trigonometric Limits

Trigonometric functions often appear in calculus problems involving limits, owing to their periodic nature and other properties. When evaluating limits of trigonometric functions like \(\sec(x)\) and \(\tan(y)\), specific angles lead to simpler calculations.For instance, evaluating \(\sec(0)\) is straightforward because \(\cos(0) = 1\), leading to \(\sec(0) = \frac{1}{1} = 1\). Similarly, \(\tan(\frac{\pi}{4}) = 1\) due to the tangent's identity at that angle. These values simplify the process of finding limits as they allow easy substitution without complex calculations. In more complicated functions, recognizing these trigonometric identities can significantly streamline solving limits by providing a direct route to an answer. This awareness of the trigonometric limits helps in computing not only multivariable functions, but also single-variable trigonometric limits.

Evaluating Limits

The evaluation of limits usually involves substituting the approaching values into the function. If direct substitution leads to an undefined situation, other techniques might be necessary like identity transformations or L'Hôpital's rule. However, in straightforward problems, as in our exercise, substitution is often sufficient.For our limit problem, we directly replaced \(x\) and \(y\) with their limiting values, \(0\) and \(\frac{\pi}{4}\), respectively. First, computing the numerator \(\sec(0) + 2\) to get \(3\), then the denominator \(3x - \tan(y)\) to receive \(-1\). Finally, combining these gives \(\frac{3}{-1} = -3\), signifying that the function reliably approaches \(-3\) as \((x, y)\) nears \((0, \frac{\pi}{4})\).Evaluating limits can be intuitive if you follow logical steps such as substitution wherever possible. Understanding the process thoroughly helps in mastering more challenging problems where intricate calculus techniques are needed.