\(\lim _{(x, y) \rightarrow(0,0)}\left(\frac{x^2-y^2}{x^2+y^2}\right) \)
will not exist
Understanding Multivariable Limits
When we evaluate the limit of a function of two variables, like \(f(x, y)\), as \((x, y)\) approaches a point \((a, b)\), the limit exists only if the function approaches the same value regardless of the path taken to reach \((a, b)\). In this question, we need to evaluate the multivariable limit as \((x, y) \rightarrow (0,0)\) for the function \(f(x, y) = \frac{x^2-y^2}{x^2+y^2}\).
Checking Different Paths for the Multivariable Limit
To determine if the multivariable limit exists, we can test the limit along different paths that approach the point \((0,0)\). If we find even two paths that yield different limit values, then the overall limit does not exist.
Path 1: Approaching along the x-axis
Let's approach \((0,0)\) along the x-axis. On the x-axis, \(y=0\). For any point \((x,y)\) on the x-axis approaching \((0,0)\) (but not equal to \((0,0)\)), we have \(x \neq 0\) and \(y=0\). Substituting \(y=0\) into the function:
\(\lim_{(x, y) \rightarrow(0,0) \text{ along } y=0} \left(\frac{x^2-y^2}{x^2+y^2}\right) = \lim_{x \rightarrow 0} \left(\frac{x^2-0^2}{x^2+0^2}\right)\)
This simplifies to:
\(\lim_{x \rightarrow 0} \left(\frac{x^2}{x^2}\right) = \lim_{x \rightarrow 0} (1)\)
The limit along the x-axis is 1.
Path 2: Approaching along the y-axis
Now, let's approach \((0,0)\) along the y-axis. On the y-axis, \(x=0\). For any point \((x,y)\) on the y-axis approaching \((0,0)\) (but not equal to \((0,0)\)), we have \(x=0\) and \(y \neq 0\). Substituting \(x=0\) into the function:
\(\lim_{(x, y) \rightarrow(0,0) \text{ along } x=0} \left(\frac{x^2-y^2}{x^2+y^2}\right) = \lim_{y \rightarrow 0} \left(\frac{0^2-y^2}{0^2+y^2}\right)\)
This simplifies to:
\(\lim_{y \rightarrow 0} \left(\frac{-y^2}{y^2}\right) = \lim_{y \rightarrow 0} (-1)\)
The limit along the y-axis is -1.
Conclusion on the Existence of the Multivariable Limit
We have found that the limit of the function approaches 1 along the x-axis and -1 along the y-axis. Since the limit values along these two different paths approaching \((0,0)\) are not equal (1 \(\neq\) -1), the multivariable limit of the function \(\frac{x^2-y^2}{x^2+y^2}\) as \((x, y) \rightarrow (0,0)\) does not exist.
This demonstrates the concept of path dependence for multivariable limits at points where the function might be undefined or have a discontinuity.
Important Questions from Functions of Several Variables
Find the simultaneous limit of function y sin(1/x) ?
Define
\(f(x, y)=\left\{\begin{array}{l} \frac{x^2-y^2}{x^2+y^2} \text { for }(x, y) \neq(0,0) \\ 0 \text { for }(x, y)=(0,0) \end{array} .\right.\)
Which of the following statements are true?Consider the function f ∶ ℝ2 → ℝ defined by
f(x, y) = x2 − y3.
Which of the following statements are true?
Let f ∶ [0,1]2 → ℝ be a function defined by
f(x, y) = \(\frac{xy}{x^2+y^2}\) if either x ≠ 0 or y ≠ 0
= 0 if x = y = 0.
Then which of the following statements are true?
Let f ∶ \(\mathbb{R}\)2 → \(\mathbb{R}\) be defined by f(x, y) = \(\begin{cases}\frac{2 x y}{x^2+y^2}, & (x, y) \neq(0,0) \\ 0, & (x, y)=(0,0) .\end{cases}\)
Define g(x, y) = \(\sum_{n=1}^{\infty} \frac{f((x-n),(y-n))}{2^n}\).
Which of the following statements are true?
