Problem 68
Question
Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths to (0,0) Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{x-y}{\sqrt{x^{2}+y^{2}}}$$
Step-by-Step Solution
Verified Answer
If yes, find its value. If no, explain why.
\(\lim_{(x,y) \to (0,0)} \frac{x - y}{\sqrt{x^{2} + y^{2}}}\)
Answer: The limit does not exist as it depends on different paths when approaching (0, 0) in polar coordinates.
1Step 1: Convert to Polar Coordinates
To convert the limit expression to polar coordinates, recall that \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). The expression in the limit becomes:
$$\frac{r\cos(\theta) - r\sin(\theta)}{\sqrt{{(r\cos(\theta))}^{2} + {(r\sin(\theta))}^{2}}}$$
2Step 2: Simplify the Expression
Begin simplifying the given expression:
$$\frac{r\cos(\theta) - r\sin(\theta)}{\sqrt{r^{2}\cos^{2}(\theta) + r^{2}\sin^{2}(\theta)}}$$
Factor out r from the numerator and the inside of the square root:
$$\frac{r(\cos(\theta) - \sin(\theta))}{\sqrt{r^{2}(\cos^{2}(\theta) + \sin^{2}(\theta))}}$$
Now, since \(\cos^{2}(\theta) + \sin^{2}(\theta) = 1\), the expression becomes:
$$\frac{r(\cos(\theta) - \sin(\theta))}{\sqrt{r^{2}\cdot 1}}$$
Which simplifies to:
$$\frac{r(\cos(\theta) - \sin(\theta))}{\abs{r}}$$
3Step 3: Take the Limit with Respect to r
Now, take the limit of the expression as \(r\rightarrow 0\):
$$\lim_{r \rightarrow 0} \frac{r(\cos(\theta) - \sin(\theta))}{\abs{r}}$$
Since the denominator will be equal to r for \(r > 0\) and \(-r\) for \(r < 0\), the limit does not exist for all paths. Therefore, the limit does not exist.
Key Concepts
Polar CoordinatesPolar ConversionPath Independence
Polar Coordinates
Polar coordinates provide a way to describe points in a plane using a radius and an angle. Unlike Cartesian coordinates, which use \(x\) and \(y\) axes, polar coordinates represent a point as \((r, \theta)\). Here, \(r\) is the distance from the origin, and \(\theta\) is the angle from the positive \(x\)-axis. This system is particularly useful when dealing with problems involving rotation or when points are naturally aligned around a central point.
To convert between Cartesian and polar coordinates, use the formulas:
To convert between Cartesian and polar coordinates, use the formulas:
- \(x = r \cos(\theta)\)
- \(y = r \sin(\theta)\)
Polar Conversion
Polar conversion involves changing expressions from Cartesian to polar form. This technique is essential when dealing with limits or integrals where symmetry around the origin simplifies analysis. In our example, the conversion of coordinates from \((x, y)\) to \((r, \theta)\) helps evaluate limits at the origin (0,0).
The expression \( \frac{x-y}{\sqrt{x^{2}+y^{2}}} \) changes to:
\[\frac{r(\cos(\theta) - \sin(\theta))}{\abs{r}}\]
This adjustment is possible due to the trigonometric identity \(\cos^2(\theta) + \sin^2(\theta) = 1\), showing how polar conversion streamlines the problem. By simplifying the calculations, polar coordinates often reveal underlying behavior that can be difficult to see in Cartesian form.
The expression \( \frac{x-y}{\sqrt{x^{2}+y^{2}}} \) changes to:
\[\frac{r(\cos(\theta) - \sin(\theta))}{\abs{r}}\]
This adjustment is possible due to the trigonometric identity \(\cos^2(\theta) + \sin^2(\theta) = 1\), showing how polar conversion streamlines the problem. By simplifying the calculations, polar coordinates often reveal underlying behavior that can be difficult to see in Cartesian form.
- Simplifies many integrals and limits.
- Useful in circular and radial symmetry cases.
- Makes evaluation of boundary limits at the origin easier.
Path Independence
Path independence is a crucial concept when evaluating limits using polar coordinates. It means that as \(r\) approaches zero, the result should be independent of the path taken. In simpler terms, regardless of how you approach the point (0,0), the limit should be consistent.
In the exercise, the limit did not exist because the expression
\[\lim_{r \rightarrow 0} \frac{r(\cos(\theta) - \sin(\theta))}{\abs{r}}\]
depended on \(\theta\). Different values of \(\theta\) lead to different results, showing that the path matters.
For a limit to exist at a point in two-dimensional space, it must yield the same result for every path. Path independence ensures that geometric interpretations, like contours and surfaces, behave predictably as they approach that point. This consistency is fundamental when analyzing real-world problems or theoretical constructs.
In the exercise, the limit did not exist because the expression
\[\lim_{r \rightarrow 0} \frac{r(\cos(\theta) - \sin(\theta))}{\abs{r}}\]
depended on \(\theta\). Different values of \(\theta\) lead to different results, showing that the path matters.
For a limit to exist at a point in two-dimensional space, it must yield the same result for every path. Path independence ensures that geometric interpretations, like contours and surfaces, behave predictably as they approach that point. This consistency is fundamental when analyzing real-world problems or theoretical constructs.
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