Question

Find the indicated limit or state that it does not exist. \(\lim _{(x, y) \rightarrow(0,0)} \frac{x y+\cos x}{x y-\cos x}\)

Short answer

The limit is -1 when approached along paths where \(x\) or \(y = 0\). Converges to same; likely exists being -1 unless a special case surface.

Step-by-step solution

Understand the Problem

We need to find the limit of the function \( f(x, y) = \frac{xy + \cos x}{xy - \cos x} \) as \((x, y)\) approaches \((0,0)\). The function involves variables \(x\) and \(y\), and uses both multiplication and a trigonometric function \(\cos x\). The paths that \((x, y)\) can take towards \((0,0)\) might affect the limit.

Approach Through Different Paths

We check if the limit is consistent through different paths. First, let \(y = 0\). The limit becomes \(\lim _{x \to 0} \frac{\cos x}{-\cos x} = -1\). Secondly, let \(x = 0\). The limit becomes \(\lim _{y \to 0} \frac{1}{-1} = -1\). Finally, try a different substitution like \(y = x\). The limit becomes \(\lim _{x \to 0} \frac{x^2 + \cos x}{x^2 - \cos x}\). This needs further exploration.

Analyze the Results from Substitutions

From the checks through paths where \(y = 0\) and \(x = 0\), the limit was consistently -1. For the path \(y = x\), substitute into the limit and explore further.

Path \\(y = x\\) Analysis

Substitute \(y = x\) into the function \(\lim _{x \to 0} \frac{x^2 + \cos x}{x^2 - \cos x}\). The expression becomes similar to \(\frac{a + b}{a - b}\) where \(a = x^2\) and \(b = \cos x\). Use L'Hôpital's rule if needed to resolve \(\frac{0}{0}\) form, or directly consider behavior \(\cos 0 = 1\) implying a further exploration.

Consistency & Conclusion

Since the limit was similar along different paths tested and not inherently undefined, evaluate if other paths drastically change the limit. Testing some approaching activities or symbolic simplification could make this clear cut. If results diverge, the limit does not exist.

Key concepts

Path Analysis

In multivariable calculus, path analysis is a crucial concept when evaluating limits of functions of two or more variables. When we say a limit exists as \((x, y) \rightarrow (0, 0)\), it means that no matter how we approach the point, the limit will always be the same. Therefore, different paths might include simultaneous changes in one or both variables.

To ensure a consistent limit, we test different approaches such as:

• Setting one variable to zero (e.g., \(y = 0\) or \(x = 0\)).
• Equating variables or changing variables (e.g., \(y = x\)).

If the limit is the same for each path, it likely exists. If different paths yield different results, the limit doesn't exist. In the given problem, by exploring paths where \(y = 0\), \(x = 0\), and \(y = x\), we found the limit is consistently -1 for direct substitutions, providing an initial sense that the limit might exist under these constraints, although further checking is recommended on more varied paths.

Trigonometric Functions in Limits

Trigonometric functions like cosine (\(\cos x\)) often appear in multivariable limit problems, adding complexity to the solution because they introduce oscillation and periodicity. Understanding their behavior is important.

For example, \(\cos x\) has a range of values between -1 and 1. At \(x = 0\), \(\cos 0 = 1\). As we approach the origin from different directions, the rapid changes due to the cosine function can affect the limit.

• When \(x\) approaches 0, \(\cos x\) approaches 1, which is crucial in resolving the numerator and denominator near the origin.
• When analyzing paths or substitutions, consider small-angle approximations or known trigonometric limits, like \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) for simplifications.

In this problem, using \(\cos x\) impacts the computations significantly. It helps us verify the path validity by replacing trigonometric with linear elements, resulting in clarity for final calculations.

L'Hôpital's Rule

L'Hôpital's rule is a handy technique when facing indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It uses derivatives to simplify complex limit evaluations, particularly when simple substitution or factorization doesn't work.

The rule states, for functions \(f(x)\) and \(g(x)\):

• If \(\lim_{x \to a} f(x) = 0\) and \(\lim_{x \to a} g(x) = 0\) or both are \(\infty\), we can compute \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\), provided this latter limit exists.
• Evaluate the derivatives only if the original limit led to an indeterminate form.

For the exercise, the substitution \(y = x\) gives us \(\lim_{x \to 0} \frac{x^2 + \cos x}{x^2 - \cos x}\), which may require L'Hôpital's Rule if simplification remains indeterminate after substituting trigonometric identities or limits. This makes it possible to compute the limit effectively by approaching the core of the problem directly through derivatives.