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Magazine Chapter 10: Problem 5
Find the limit, if it exists, or type \(\mathrm{N}\) if it does not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x+13 y)^{2}}{x^{2}+13^{2} y^{2}}=$$ _____________.
Short Answer
Expert verified
The limit of the given function as (x, y) approaches (0, 0) is 0.
Step by step solution
01
Check for a direct substitution
If the function is defined at the point (0,0), we can directly substitute x and y with 0 and find the limit.
However, substituting x and y with 0, the function becomes \(\frac{(0+13\cdot0)^2}{0^2 + (13\cdot0)^2} = \frac{0}{0}\), which is an indeterminate form. Thus, we need to use other methods to find the limit.
02
Polar coordinate transformation
Since we have a 2-variable limit, we can try transforming the function into polar coordinates and then find the limit by using the transformed form.
To do this, we will substitute x with \(r\cos\theta\) and y with \(r\sin\theta\), where r is the radial distance from the origin, and \(\theta\) is the angle.
Our given function becomes:
\(\frac{((r\cos\theta)+13(r\sin\theta))^2}{(r\cos\theta)^2+ (13r\sin\theta)^2}\)
After substituting, we can simplify the function:
03
Simplify the function
Simplifying the given function after the polar coordinate transformation:
\(\lim_{r \to 0} \frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\cdot r^2\)
Now, notice that the function does not depend on \(\theta\) anymore. Instead, we can remove the limits depending on r:
\(\frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\)
04
Evaluate the limit
Now, let us see if this limit exists:
\(\lim_{(x, y) \to (0,0)} \frac{(x+13 y)^{2}}{x^{2}+13^{2} y^{2}} = \lim_{r \to 0} \frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\cdot r^2 = 0\)
Since the part \(\frac{(\cos\theta + 13\sin\theta)^2}{(\cos^2\theta + 13^2\sin^2\theta)}\) approaches a constant as r approaches 0, the limit converges to 0 as the whole expression is multiplied by \(r^2\), which approaches 0 as r approaches 0.
Thus, the limit of the function as (x, y) approaches (0, 0) is 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates Transformation
When evaluating limits in two variables, polar coordinates can be a powerful tool. In this system, each point in the plane is represented by
In the problem, this substitution transformed our expression into:\[ \frac{((r\cos\theta)+13(r\sin\theta))^2}{(r\cos\theta)^2+ (13r\sin\theta)^2} \]After simplifying, we were able to isolate and evaluate the dependence on \(r\) and \(\theta\). This transformation helps us see if the limit uniformly approaches the origin from all directions.
- an angle \(\theta\)
- a radial distance \(r\) from the origin
In the problem, this substitution transformed our expression into:\[ \frac{((r\cos\theta)+13(r\sin\theta))^2}{(r\cos\theta)^2+ (13r\sin\theta)^2} \]After simplifying, we were able to isolate and evaluate the dependence on \(r\) and \(\theta\). This transformation helps us see if the limit uniformly approaches the origin from all directions.
Indeterminate Form
An indeterminate form occurs when direct substitution into a function leads to expressions like \(\frac{0}{0}\). These forms don't provide clear information about a limit's behavior. In the problem at hand, substituting \((x, y) = (0, 0)\) directly yielded the indeterminate form \(\frac{0}{0}\). This suggested that we couldn't evaluate the limit directly.
Indeterminate forms often require us to apply other methods, such as:
Indeterminate forms often require us to apply other methods, such as:
- L'H么pital's Rule (for certain 1-variable functions)
- Algebraic manipulation
- Coordinate transformations (as seen in this exercise)
Multivariable Calculus
Multivariable calculus expands on principles from single-variable calculus to functions with more than one variable. It deals with topics like:
Understanding these concepts helps when navigating through complex geometries and varying directions within multivariable functions.
- Partial derivatives
- Multiple integrals
- Vector fields
- Limits involving multiple variables
Understanding these concepts helps when navigating through complex geometries and varying directions within multivariable functions.
Limit Evaluation Techniques
To evaluate limits in two variables, various techniques are employed to provide accuracy and insights. The common strategies include:
Mastering these techniques provides a robust foundation for handling limits in multivariable calculus efficiently.
- Direct substitution (if applicable)
- Path approach, where different paths towards the limit point yield consistent results
- Using a polar coordinates transformation, as seen in the problem
- Simplifying expressions to remove indeterminate forms
Mastering these techniques provides a robust foundation for handling limits in multivariable calculus efficiently.
