Evaluating The Limit Of A Multivariable Function Lim (p->(1,3,4)) (1/x + 1/y + 1/z)

Introduction

In the realm of multivariable calculus, understanding limits is fundamental. Limits form the bedrock upon which concepts like continuity, derivatives, and integrals are built. Evaluating limits of multivariable functions, however, can be more intricate than their single-variable counterparts. This article delves into the evaluation of a specific limit: $\lim _{p \rightarrow(1,3,4)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$. We will dissect the problem, explore the underlying principles, and provide a comprehensive solution. This limit problem serves as an excellent example to illustrate how to approach limits in higher dimensions and highlights the importance of understanding the behavior of functions as we approach a specific point in space.

The expression $\lim _{p \rightarrow(1,3,4)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ represents the limit of the function $f(x, y, z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ as the point $p(x, y, z)$ approaches the point (1, 3, 4) in three-dimensional space. To evaluate this limit, we need to determine the value that the function $f(x, y, z)$ approaches as $(x, y, z)$ gets arbitrarily close to (1, 3, 4). This seemingly straightforward problem opens the door to a deeper understanding of continuity and the behavior of multivariable functions. This exploration will equip you with the tools to tackle more complex limit problems and provide a solid foundation for further studies in multivariable calculus. We will not only compute the limit but also discuss why this particular limit is well-behaved and what conditions might lead to more challenging limit evaluations.

Understanding Multivariable Limits

Multivariable limits extend the concept of limits from single-variable calculus to functions of multiple variables. In single-variable calculus, we examine the behavior of a function as the input variable approaches a specific value from the left and the right. In the multivariable context, we consider the behavior of a function as the input point approaches a specific point from any direction within the function's domain. This adds a layer of complexity because there are infinitely many paths along which a point can approach another point in higher dimensions. Therefore, for a multivariable limit to exist, the function must approach the same value regardless of the path taken. If different paths yield different limit values, then the overall limit does not exist. The key here is to consider different paths of approach. If the function approaches the same value regardless of the path, the limit likely exists and equals that value. If we find even two paths that yield different limits, we can definitively conclude that the limit does not exist. This path-dependent behavior is a crucial difference between single-variable and multivariable limits. This concept can be visualized as approaching a mountain peak. No matter which trail you take up the mountain, you should reach the same summit (limit). If different trails lead to different heights, there's no single peak (limit).

The Direct Substitution Method

The direct substitution method is often the first approach to try when evaluating limits, both in single-variable and multivariable calculus. This method involves directly substituting the values that the variables are approaching into the function. If the result is a finite number, then the limit is simply that number. However, direct substitution can sometimes lead to indeterminate forms such as 0/0 or ∞/∞. In such cases, other techniques, such as factoring, rationalizing, or L'Hôpital's Rule (in single-variable calculus), may be necessary. In the context of multivariable limits, if direct substitution yields a finite value, it strongly suggests that the limit exists and is equal to that value, especially if the function is continuous at the point in question. However, it's important to remember that direct substitution only provides a conclusive answer when it results in a finite value. If it leads to an indeterminate form or an undefined expression, further investigation is required. Direct substitution hinges on the idea of continuity. If a function is continuous at a point, the limit as we approach that point is simply the function's value at that point. This is what allows direct substitution to work. However, we must always be mindful of potential discontinuities or indeterminate forms.

Evaluating the Limit

To evaluate the limit $\lim _{p \rightarrow(1,3,4)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$, where $p = (x, y, z)$, we can use the direct substitution method. This method is applicable because the function $f(x, y, z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is a sum of rational functions, and each of these rational functions is continuous everywhere except where its denominator is zero. Since the point (1, 3, 4) does not make any of the denominators zero, the function is continuous at this point. Therefore, we can directly substitute the values $x = 1$, $y = 3$, and $z = 4$ into the function.

Substituting these values, we get:

$f(1, 3, 4) = \frac{1}{1} + \frac{1}{3} + \frac{1}{4}$

To add these fractions, we need to find a common denominator. The least common multiple of 1, 3, and 4 is 12. So, we rewrite the fractions with a denominator of 12:

$f(1, 3, 4) = \frac{12}{12} + \frac{4}{12} + \frac{3}{12}$

Now, we can add the numerators:

$f(1, 3, 4) = \frac{12 + 4 + 3}{12} = \frac{19}{12}$

Since we obtained a finite value by direct substitution, we can conclude that the limit exists and is equal to this value. This direct approach works because the function is well-behaved (continuous) at the point we're approaching. If the function had discontinuities or resulted in an indeterminate form (like 0/0) upon substitution, we would need to employ more advanced techniques to evaluate the limit. However, in this case, the direct method provides a straightforward and correct solution. The absence of any singularities or discontinuities near the point (1,3,4) makes direct substitution a reliable method here. Had there been any such issues, a more nuanced approach considering paths of approach would have been necessary.

Continuity and Direct Substitution

The success of the direct substitution method hinges on the concept of continuity. A function is continuous at a point if the limit of the function as the input approaches that point is equal to the function's value at the point. In other words, if $f(x, y, z)$ is continuous at $(a, b, c)$, then:

$\lim_{(x, y, z) \rightarrow (a, b, c)} f(x, y, z) = f(a, b, c)$

In our case, the function $f(x, y, z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ is continuous at the point (1, 3, 4) because each term in the sum is a rational function, and rational functions are continuous everywhere except where their denominators are zero. Since none of the denominators (x, y, or z) are zero at (1, 3, 4), the function is continuous there. This continuity allows us to confidently use direct substitution to evaluate the limit. The concept of continuity is crucial in limit evaluation because it allows us to bypass potentially complex limit calculations in many cases. If we know a function is continuous at the point we're interested in, we can simply evaluate the function at that point to find the limit. However, it's crucial to verify continuity before applying direct substitution, as this method is not valid for discontinuous functions at the point of interest. Understanding the conditions for continuity is therefore essential for efficient and accurate limit calculations.

Conclusion

The limit $\lim _{p \rightarrow(1,3,4)}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ can be evaluated using the direct substitution method due to the continuity of the function $f(x, y, z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ at the point (1, 3, 4). By substituting the values $x = 1$, $y = 3$, and $z = 4$ into the function, we find that the limit is $\frac{19}{12}$. This example illustrates the importance of understanding continuity in the context of multivariable limits. When a function is continuous at the point of interest, direct substitution provides a straightforward way to evaluate the limit. This problem, while seemingly simple, underscores a fundamental principle in multivariable calculus: the interplay between continuity and limit evaluation. The direct substitution method, when applicable, is a powerful tool for finding limits quickly and efficiently. However, it is essential to always check for continuity before applying this method to avoid potential errors. The understanding of continuity and the ability to identify situations where direct substitution is valid are crucial skills for anyone working with multivariable limits. This example serves as a solid foundation for tackling more complex limit problems in the future. The concepts explored here will be valuable as you delve deeper into multivariable calculus and encounter more challenging functions and limit scenarios.