Problem 12
Question
Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(-1,-2)} \frac{x^{2}-y^{2}}{2 x y+2}\)
Step-by-Step Solution
Verified Answer
The limit is
\(-\frac{1}{2}\).
1Step 1: Identify the Form of the Limit
We need to check if the limit \[\lim _{(x, y) \rightarrow(-1,-2)} \frac{x^{2}-y^{2}}{2xy+2}\]is an indeterminate form. Substituting \(x = -1\) and \(y = -2\), we get:\[\frac{(-1)^2 - (-2)^2}{2(-1)(-2) + 2} = \frac{1 - 4}{4 + 2} = \frac{-3}{6} = -\frac{1}{2}\]This means the expression is not indeterminate, and we can directly substitute the values to find the limit.
2Step 2: Direct Substitution
Since the expression is not indeterminate, we substitute \(x = -1\) and \(y = -2\) directly into the limit expression:\[x^2 - y^2 = (-1)^2 - (-2)^2 = 1 - 4 = -3\]\[2xy + 2 = 2(-1)(-2) + 2 = 4 + 2 = 6\]Therefore, the limit simplifies to:\[\lim _{(x, y) \rightarrow(-1,-2)} \frac{-3}{6} = -\frac{1}{2}\]
Key Concepts
Indeterminate FormsDirect SubstitutionMultivariable Limits
Indeterminate Forms
In calculus, indeterminate forms occur when you try to evaluate a limit and end up with a form that does not easily point to a specific value. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \), \( 0 \times \infty \), and others.
They often appear when both the numerator and the denominator of a fraction tend toward zero. In these cases, additional algebraic manipulation or special techniques, such as L'Hôpital's Rule, are required to evaluate the limit further.
In the given exercise, we were fortunate that upon substituting \((x, y) = (-1, -2)\), the expression wasn't an indeterminate form. This allowed us to directly calculate the limit without additional techniques. Recognizing whether an expression is indeterminate shapes how you approach solving the problem. Once you identify indeterminate forms, knowing the proper techniques allows you to evaluate limits correctly.
They often appear when both the numerator and the denominator of a fraction tend toward zero. In these cases, additional algebraic manipulation or special techniques, such as L'Hôpital's Rule, are required to evaluate the limit further.
In the given exercise, we were fortunate that upon substituting \((x, y) = (-1, -2)\), the expression wasn't an indeterminate form. This allowed us to directly calculate the limit without additional techniques. Recognizing whether an expression is indeterminate shapes how you approach solving the problem. Once you identify indeterminate forms, knowing the proper techniques allows you to evaluate limits correctly.
Direct Substitution
Direct substitution is one of the simplest methods to evaluate a limit. When the limit expression is not indeterminate, direct substitution allows us to substitute the value toward which the variable tends directly into the equation.
This method is simple but powerful. If substituting values into the expression gives a well-defined number, you've successfully found the limit. As seen in the step-by-step solution of our exercise, substituting \((x, y) = (-1, -2)\) resulted in a straightforward computation: \(\frac{-3}{6} = -\frac{1}{2}\).
This approach effectively verified that the expression simplifies neatly without complications, making it suitable for easy computation. Always remember: direct substitution is your best first attempt at finding a limit if the expression is defined and non-indeterminate.
This method is simple but powerful. If substituting values into the expression gives a well-defined number, you've successfully found the limit. As seen in the step-by-step solution of our exercise, substituting \((x, y) = (-1, -2)\) resulted in a straightforward computation: \(\frac{-3}{6} = -\frac{1}{2}\).
This approach effectively verified that the expression simplifies neatly without complications, making it suitable for easy computation. Always remember: direct substitution is your best first attempt at finding a limit if the expression is defined and non-indeterminate.
Multivariable Limits
Multivariable limits involve functions of two or more variables, offering a more nuanced exploration compared to single-variable limits. These limits consider how expressions change as the variables approach specific values simultaneously.
To evaluate a multivariable limit, the process is quite similar to evaluating single-variable limits, but with an additional dimension in the variable space. You are determining how the function behaves as the variables approach specific points from all directions in their multidimensional space.
In our example, we assessed the limit \(\lim _{(x, y) \rightarrow(-1,-2)} \frac{x^{2}-y^{2}}{2xy+2}\). We evaluated it by substituting values directly, because the form was suitable for such treatment. However, for complex multivariable limits not conducive for simple substitution, approaching from different paths or using polar coordinates can sometimes provide further insight.
Understanding multivariable limits is crucial for analyzing real-world phenomena where multiple factors simultaneously change and influence outcomes.
To evaluate a multivariable limit, the process is quite similar to evaluating single-variable limits, but with an additional dimension in the variable space. You are determining how the function behaves as the variables approach specific points from all directions in their multidimensional space.
In our example, we assessed the limit \(\lim _{(x, y) \rightarrow(-1,-2)} \frac{x^{2}-y^{2}}{2xy+2}\). We evaluated it by substituting values directly, because the form was suitable for such treatment. However, for complex multivariable limits not conducive for simple substitution, approaching from different paths or using polar coordinates can sometimes provide further insight.
Understanding multivariable limits is crucial for analyzing real-world phenomena where multiple factors simultaneously change and influence outcomes.
Other exercises in this chapter
Problem 11
Evaluate the negative binomial model for the first 25 generations when \(a=0.02, c=3, k=0.75\), and \(b=1.5\). For the initial host density, choose \(N_{0}=100\
View solution Problem 12
Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=e^{-y^{2}} \sin \left(x^{2}+y^{2}\right)\)
View solution Problem 12
Evaluate each function at the given point. \(f(x, y)=\frac{x^{2}}{y}\) at \((3,2)\)
View solution Problem 12
Show that \(f(x, y)\) is differentiable at the indicated point. \(f(x, y)=x y+3 x^{2} ;(1,0)\)
View solution