Problem 28 - Single Variable Calculus

Question

You are given that \(\lim _{x \rightarrow a} f(x)=2\), \(\lim _{x \rightarrow a a} g(x)=4\), and \(\lim _{x \rightarrow a} h(x)=-1 .\) Find the indicated limit. \(\lim _{x \rightarrow a} \frac{\sqrt[3]{f(x) g(x)}}{\sqrt{f(x) g(x)+1}}\)

Step-by-Step Solution

Answer

The short answer to the problem is: \(\lim_{x \rightarrow a} \frac{\sqrt[3]{f(x) g(x)}}{\sqrt{f(x) g(x)+1}} = {\sqrt[3]{2}} \cdot 4^{\frac{1}{3}} \cdot \frac{1}{3}\).

Step 1: Apply the limit properties

We can rewrite the expression inside the limit as follows: \[\lim_{x \rightarrow a} \frac{\sqrt[3]{f(x)g(x)}}{\sqrt{f(x)g(x)+1}} = \lim_{x \rightarrow a} {\sqrt[3]{f(x)}} \cdot \lim_{x \rightarrow a} {\sqrt[3]{g(x)}} \cdot \frac{1}{\sqrt{\lim_{x \rightarrow a} {f(x)g(x)+1}}}\]

Step 2: Substitute the given limits

We can now substitute the given limits for \(f(x)\), \(g(x)\), and \(h(x)\) into the expression: \[\lim_{x \rightarrow a} {\sqrt[3]{f(x)}} \cdot \lim_{x \rightarrow a} {\sqrt[3]{g(x)}} \cdot \frac{1}{\sqrt{\lim_{x \rightarrow a} {f(x)g(x)+1}}} = {\sqrt[3]{2}} \cdot {\sqrt[3]{4}} \cdot \frac{1}{\sqrt{2 \cdot 4+1}}\]

Step 3: Simplify the expression

Now we simplify the expression: \[{\sqrt[3]{2}} \cdot {\sqrt[3]{4}} \cdot \frac{1}{\sqrt{2 \cdot 4+1}} = {\sqrt[3]{2}} \cdot {\sqrt[3]{4}} \cdot \frac{1}{\sqrt{9}}\] Since \(\sqrt[3]{4} = 4^{\frac{1}{3}}\) and \(\sqrt{9} = 3\), the final expression is: \[{\sqrt[3]{2}} \cdot 4^{\frac{1}{3}} \cdot \frac{1}{3}\] So, the limit of the given expression as \(x\) approaches \(a\) is: \[\lim_{x \rightarrow a} \frac{\sqrt[3]{f(x) g(x)}}{\sqrt{f(x) g(x)+1}} = {\sqrt[3]{2}} \cdot 4^{\frac{1}{3}} \cdot \frac{1}{3}\]

Key Concepts

Limit Properties

When it comes to calculus, one of the foundational concepts is that of limits. Limits describe how a function behaves as the input approaches a certain value. But what makes limits especially powerful are the properties they have which allow us to manipulate and evaluate them in various ways.

Some of the key properties of limits include the sum, difference, product, and quotient of limits. For instance, if we know the limit of two functions as the input approaches the same value, we can combine these limits to find the limit of their sum, difference, or product. These properties are not just handy rules; they reflect the predictable behavior of limits under basic arithmetic operations.

Specifically, if we have two functions, say, f(x) and g(x), with their limits as x approaches a being known, the limit of their sum is the sum of their limits. The same principle applies to differences and products. However, caution is needed with quotients: the limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.

In our given exercise, we utilized these properties to break down the original limit into more manageable parts. This way, we can find the limit of each part separately and then combine them to find the limit of the complex expression.

Limits of Functions

Understanding the behavior of functions as their input approaches a particular value is one of the goals of studying limits in calculus. This ability to predict the function's behavior is central to the concept of continuity and to the calculation of derivatives later on.

When we discuss 'limits of functions,' we are typically trying to answer the question: what is the function's output as the input gets arbitrarily close to a specific number? For example, as x approaches some value a, what does f(x) approach?

To showcase this, in the textbook example, we see how limits are determined separately for the functions f(x), g(x), and h(x) as x approaches a value a, and then how these individual limits contribute to the limit of a more complicated function. This segmentation simplifies complex problems, making them more manageable - a key skill in calculus. It’s also a great example of how understanding the limit of simple functions can provide insight into the behavior of more complex expressions constructed from these functions.

Rationalizing Limits

When faced with a limit involving a radical—whether it's a square root, cube root, or another root—things can get tricky. This is where the technique of rationalizing comes into play. The goal of rationalizing is to eliminate the radical from the denominator of a fraction or from under the radical sign to simplify the limit process.

Rationalizing is particularly useful when the limit involves a complex expression with radicals because it transforms the expression into one that is easier to manage and more straightforward to evaluate. We typically accomplish this by multiplying the numerator and denominator by a 'conjugate'—a similar expression but with an opposite sign between its terms—or by some variant of the original radical that will help cancel out the radical parts once expanded.

In our exercise, there is no radical in the denominator, but we do see a cube root in the numerator. If a radical was present in the denominator, we would rationalize it for simplification. This technique didn’t come into play in this particular problem, but it's an important strategy to be aware of for handling other limits that involve roots, providing a pathway to reach a limit that initially seems indeterminate or complex.