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Magazine Chapter 15: Problem 69
Use the method of your choice to evaluate the following limits. $$\lim _{(x, y) \rightarrow(2,0)} \frac{1-\cos y}{x y^{2}}$$
Short Answer
Expert verified
Answer: The limit of the expression is 1.
Step by step solution
01
Identify the given expression
We are asked to find the limit of the following expression as \((x, y) \rightarrow (2, 0)\):
$$f(x, y) = \frac{1-\cos y}{x y^{2}}$$
02
Simplify the expression using trigonometric identity
Recall the half-angle identity: \(\sin^2(\frac{y}{2}) = \frac{1-\cos y}{2}\). Using this identity, we can rewrite the given expression as:
$$f(x, y) = \frac{2\sin^2(\frac{y}{2})}{x y^{2}}$$
03
Apply limit properties
We want to find the limit as \((x, y) \rightarrow (2, 0)\). Since the expression for \(f(x, y)\) contains a combination of variables \(x\) and \(y\), we can split it into separate parts:
$$\lim_{(x, y) \rightarrow (2, 0)} f(x, y) = \lim_{y \rightarrow 0} \left(\frac{2 \sin^2(\frac{y}{2})}{y^2}\right) \cdot \lim_{x \rightarrow 2} \left(\frac{1}{x}\right)$$
04
Evaluate limits separately
First, we evaluate the limit as \(y \rightarrow 0\) using L'H么pital's rule:
$$\lim_{y \rightarrow 0} \frac{2 \sin^2(\frac{y}{2})}{y^2}$$
As \(\lim_{y \rightarrow 0} \frac{2 \sin^2(\frac{y}{2})}{y^2}\) has indeterminate form \(\frac{0}{0}\), we can apply L'H么pital's rule:
$$\lim_{y \rightarrow 0} \frac{2 \sin^2(\frac{y}{2})}{y^2} = \lim_{y \rightarrow 0} \frac{4 \sin(\frac{y}{2})\cos(\frac{y}{2})}{2y}$$
The new expression is again in indeterminate form \(\frac{0}{0}\), Apply L'H么pital's rule once more:
$$\lim_{y \rightarrow 0} \frac{4 \sin(\frac{y}{2})\cos(\frac{y}{2})}{2y} = \lim_{y \rightarrow 0} \frac{4 \cos^2(\frac{y}{2}) - 4 \sin^2(\frac{y}{2})}{2}$$
As there is no more indeterminate form when \(y \rightarrow 0\):
$$\lim_{y \rightarrow 0} \frac{4 \cos^2(\frac{y}{2}) - 4 \sin^2(\frac{y}{2})}{2} = \frac{4 - 0}{2} = 2$$
Now we can evaluate the second limit as \(x \rightarrow 2\):
$$\lim_{x \rightarrow 2} \frac{1}{x} = \frac{1}{2}$$
05
Combine the limits to get the final result
Now, we multiply the two limits that we just evaluated:
$$\lim_{(x, y) \rightarrow (2, 0)} f(x, y) = \left(\lim_{y \rightarrow 0} \frac{2 \sin^2(\frac{y}{2})}{y^2}\right) \cdot \left(\lim_{x \rightarrow 2} \frac{1}{x}\right) = 2 \cdot \frac{1}{2} = 1$$
So the limit of the given expression as \((x, y) \rightarrow (2, 0)\) is \(\boxed{1}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Limits
Trigonometric limits often arise in calculus when dealing with functions near certain points. These involve limits of trigonometric functions like sine, cosine, tangent, and others. One common identity used in these types of limit problems is the half-angle identity. In this exercise, the half-angle identity \( \sin^2\left(\frac{y}{2}\right) = \frac{1 - \cos y}{2} \) was applied to transform the given expression into a more workable form.
This technique simplifies the trigonometric expression and makes it easier to handle especially if it results in an indeterminate form like \( \frac{0}{0} \). By using trigonometric identities, you'll often simplify limits that would otherwise be complex to decipher.
When solving trigonometric limits, always look for opportunities to use these identities to reduce the complexity of expressions. This is a valuable skill, particularly for multivariable functions where one variable's behavior must be isolated to evaluate the limit properly.
This technique simplifies the trigonometric expression and makes it easier to handle especially if it results in an indeterminate form like \( \frac{0}{0} \). By using trigonometric identities, you'll often simplify limits that would otherwise be complex to decipher.
When solving trigonometric limits, always look for opportunities to use these identities to reduce the complexity of expressions. This is a valuable skill, particularly for multivariable functions where one variable's behavior must be isolated to evaluate the limit properly.
L'H么pital's Rule
L'H么pital's Rule is an essential tool in calculus used to evaluate limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule states that if the limit of a quotient results in one of these forms, you can take the derivatives of the numerator and the denominator separately and then re-evaluate the limit.
In this exercise, L'H么pital's Rule was applied twice. The original expression after using the trigonometric identity still resulted in the indeterminate form \( \frac{0}{0} \). By differentiating the numerator and the denominator, a new expression emerged. However, this too was \( \frac{0}{0} \), prompting another application of the rule.
The crucial part of applying L'H么pital's Rule is ensuring that the conditions for the rule are met, specifically that the original limit is indeed an indeterminate form. Once you attain a determinate limit through differentiation, you can stop applying the rule and solve the expression directly.
In this exercise, L'H么pital's Rule was applied twice. The original expression after using the trigonometric identity still resulted in the indeterminate form \( \frac{0}{0} \). By differentiating the numerator and the denominator, a new expression emerged. However, this too was \( \frac{0}{0} \), prompting another application of the rule.
The crucial part of applying L'H么pital's Rule is ensuring that the conditions for the rule are met, specifically that the original limit is indeed an indeterminate form. Once you attain a determinate limit through differentiation, you can stop applying the rule and solve the expression directly.
Indeterminate Forms
Indeterminate forms occur in calculus when a limit leads to expressions where the behavior is not immediately clear. The most common indeterminate forms are \( \frac{0}{0} \), \( 0 \cdot \infty \), \( \infty - \infty \), and others. These forms suggest that straightforward substitution will not solve the limit, and thus techniques such as algebraic manipulation, trigonometric identities, or L'H么pital's Rule may be utilized.
In this limit problem, the first sight of an indeterminate form \( \frac{0}{0} \) was seen when applying the limit \( y \rightarrow 0 \) using the simplified trigonometric expression. Rather than directly solving it, further simplifications and derivative applications (like those enabled by L'H么pital's Rule) helped reform the problematic expression into something manageable.
Understanding indeterminate forms is crucial because they often disguise a hidden limit that can be resolved with the right approach. Recognizing when a limit is in such a form is the first step in choosing the appropriate strategy to evaluate it effectively.
In this limit problem, the first sight of an indeterminate form \( \frac{0}{0} \) was seen when applying the limit \( y \rightarrow 0 \) using the simplified trigonometric expression. Rather than directly solving it, further simplifications and derivative applications (like those enabled by L'H么pital's Rule) helped reform the problematic expression into something manageable.
Understanding indeterminate forms is crucial because they often disguise a hidden limit that can be resolved with the right approach. Recognizing when a limit is in such a form is the first step in choosing the appropriate strategy to evaluate it effectively.
