Question
The limit of the function f (x, y) = x + y - 6 at x = 1; y = 2 is ?
The correct answer is
-3
Understanding the Limit of a Two-Variable Function
The problem asks us to find the limit of the function \( f(x, y) = x + y - 6 \) as \( x \) approaches 1 and \( y \) approaches 2. This is a question about evaluating the limit of a function of two variables at a specific point in its domain. A function of two variables, like \( f(x, y) = x + y - 6 \), is considered continuous at a point \((a, b)\) if the limit of the function as \((x, y)\) approaches \((a, b)\) is equal to the function's value at \((a, b)\). That is, \( \lim_{(x,y) \to (a,b)} f(x, y) = f(a, b) \). The given function \( f(x, y) = x + y - 6 \) is a polynomial in two variables (\( x \) and \( y \)). Polynomial functions are known to be continuous everywhere in their domain. The domain of this function is all pairs of real numbers \((x, y)\). Since the function is continuous at every point, including \((1, 2)\), we can find the limit by simply evaluating the function at the given point \((x, y) = (1, 2)\).Step-by-Step Limit Calculation
To find the limit of \( f(x, y) = x + y - 6 \) at \( x = 1 \) and \( y = 2 \), we substitute these values directly into the function.- Identify the function: \( f(x, y) = x + y - 6 \)
- Identify the point: The point is \((x, y) = (1, 2)\).
- Substitute the values into the function: \( f(1, 2) = (1) + (2) - 6 \)
- Perform the arithmetic: \( f(1, 2) = 3 - 6 \)
- Calculate the result: \( f(1, 2) = -3 \)
Final Answer for the Limit
The limit of the function \( f(x, y) = x + y - 6 \) at \( x = 1 \) and \( y = 2 \) is -3.| Term | Description |
|---|---|
| Function | A rule that assigns to each input pair (x, y) exactly one output value f(x, y). |
| Limit | The value that a function approaches as the input approaches a specific point. |
| Continuity | A property of a function where small changes in the input result in small changes in the output; a function is continuous at a point if its limit at that point equals its value at that point. |
| Two-Variable Function | A function whose input consists of two independent variables, typically denoted as f(x, y). |
Revision Table: Key Concepts
Reviewing the essential terms helps in understanding the concept of limits for multivariable functions.| Concept | Brief Explanation |
|---|---|
| Limit Evaluation for Continuous Functions | For a continuous function f(x, y) at a point (a, b), the limit as (x, y) approaches (a, b) is simply f(a, b). |
| Polynomials in Two Variables | Functions formed by adding, subtracting, and multiplying variables x and y and constants (e.g., x + y - 6, x2y, 5x - 2y3). These are continuous everywhere. |
| Approaching a Point (a, b) | Means that the point (x, y) gets arbitrarily close to (a, b). For limits to exist in multivariable calculus, the function must approach the same value regardless of the path taken towards the point (a, b). |
Additional Information on Multivariable Limits
While evaluating the limit by direct substitution works for continuous functions like the one in this problem, finding limits for other types of multivariable functions can be more complex.- For functions that are not continuous or are undefined at the point \((a, b)\), direct substitution is not possible.
- In such cases, one must investigate if the function approaches the same value along all possible paths towards \((a, b)\). If the function approaches different values along different paths, the limit does not exist.
- Methods for finding multivariable limits include polar coordinates, changing variables, or algebraic manipulation (like factoring or multiplying by the conjugate).
- The concept of limits is fundamental to defining continuity, derivatives, and integrals for functions of multiple variables.
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Important Questions from Limits
The value of \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{3x - 6}}\) is:
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- \(\mathop {\lim }\limits_{x \to - 5} \frac{{\sqrt {\left( {2x + 35} \right)} - 5}}{{x + 5}}\)
The value of \(\mathop {\lim }\limits_{x \to 0} \frac{{{x^3} - {\rm{sin}}\left( x \right)}}{x}\;\)
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