Problem 24
Question
Assume \(\lim _{x \rightarrow 1} f(x)=8, \lim _{x \rightarrow 1} g(x)=3,\) and \(\lim _{x \rightarrow 1} h(x)=2 .\) Compute the following limits and state the limit laws used to justify your computations. \(\lim _{x \rightarrow 1} \sqrt[3]{f(x) g(x)+3}\)
Step-by-Step Solution
Verified Answer
Answer: The limit of \(\lim_{x \rightarrow 1} \sqrt[3]{f(x) \cdot g(x) + 3}\) is 3. We used the following limit laws during the computation:
1. Product of Limits
2. Sum of Limits
3. Constant multiple
4. Power of Limit
1Step 1: Apply the product of limits law
Using the product of limits law, we know that \(\lim_{x \rightarrow a} [f(x) \cdot g(x)] = \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x)\). So, first we need to compute the limit of the product of \(f(x)\) and \(g(x)\) as \(x \rightarrow 1\).
\(\lim _{x \rightarrow 1} [f(x) \cdot g(x)] = \lim_{x \rightarrow 1} f(x) \cdot \lim_{x \rightarrow 1} g(x) = 8 \cdot 3 = 24\)
2Step 2: Apply the sum of limits law and constant multiple
Next, we have to add a constant 3 to the previous result. The constant multiple and sum of limits can be applied as follows.
\(\lim_{x \rightarrow 1} [f(x) \cdot g(x) + 3] = \lim _{x \rightarrow 1} [f(x) \cdot g(x)] + \lim_{x \rightarrow 1} 3 = 24 + 3 = 27\)
3Step 3: Apply the power of limit law
Finally, we apply the power of limit law to compute the cube root. Let the function \(k(x) = f(x) \cdot g(x) + 3\)
So, \(\lim_{x \rightarrow 1} \sqrt[3]{k(x)} = \sqrt[3]{\lim_{x \rightarrow 1} k(x)} = \sqrt[3]{27} = 3\)
The limit of \(\lim_{x \rightarrow 1} \sqrt[3]{f(x) \cdot g(x) + 3}\) is 3, and we applied the following limit laws to justify our computations:
1. Product of Limits
2. Sum of Limits
3. Constant multiple
4. Power of Limit
Key Concepts
Product of Limits LawSum of Limits LawConstant Multiple in LimitsPower of Limit Law
Product of Limits Law
When studying calculus, the product of limits law is a fundamental concept that allows the simplification of limit calculations involving multiplication. In essence, this law states that if you have two functions, let's call them f(x) and g(x), the limit of their product as x approaches a certain value is simply the product of their individual limits at that value. Mathematically, we express this as:
\[\begin{equation}\lim_{{x \to a}} [f(x) \cdot g(x)] = \lim_{{x \to a}} f(x) \cdot \lim_{{x \to a}} g(x)\end{equation}\]
This is incredibly useful because it breaks down a potentially complicated product into smaller, more manageable pieces. However, it’s vital to ensure that the individual limits exist and are finite because the product law is valid under these conditions. In the given exercise, since both functions f(x) and g(x) approach finite limits as x approaches 1, their product also approaches a specific limit, which is the product of the two separate limits.
\[\begin{equation}\lim_{{x \to a}} [f(x) \cdot g(x)] = \lim_{{x \to a}} f(x) \cdot \lim_{{x \to a}} g(x)\end{equation}\]
This is incredibly useful because it breaks down a potentially complicated product into smaller, more manageable pieces. However, it’s vital to ensure that the individual limits exist and are finite because the product law is valid under these conditions. In the given exercise, since both functions f(x) and g(x) approach finite limits as x approaches 1, their product also approaches a specific limit, which is the product of the two separate limits.
Sum of Limits Law
Parallel to multiplying functions, we can also add them, and this is where the sum of limits law comes into play. According to this principle:
\[\begin{equation}\lim_{{x \to a}} [f(x) + g(x)] = \lim_{{x \to a}} f(x) + \lim_{{x \to a}} g(x)\end{equation}\]
Like its multiplicative counterpart, the sum of limits law permits us to take the limit of each function separately and then add them together. It is a convenient way to handle complicated sums, simplifying the calculation process. Similar to the product law, it requires that the individual limits of f(x) and g(x) exist. The exercise demonstrates this law in action, by allowing the addition of a constant to the product of the limits effortlessly.
\[\begin{equation}\lim_{{x \to a}} [f(x) + g(x)] = \lim_{{x \to a}} f(x) + \lim_{{x \to a}} g(x)\end{equation}\]
Like its multiplicative counterpart, the sum of limits law permits us to take the limit of each function separately and then add them together. It is a convenient way to handle complicated sums, simplifying the calculation process. Similar to the product law, it requires that the individual limits of f(x) and g(x) exist. The exercise demonstrates this law in action, by allowing the addition of a constant to the product of the limits effortlessly.
Constant Multiple in Limits
Limits in calculus respect scalar multiplication, which is described by the constant multiple in limits rule. This law can be stated as follows: If you have a limit of a function f(x), and you multiply that function by a constant c, the limit of the product is the constant multiplied by the limit of the function, which can be algebraically penned as:
\[\begin{equation}\lim_{{x \to a}} [c \cdot f(x)] = c \cdot \lim_{{x \to a}} f(x)\end{equation}\]
What makes this rule so helpful is its allowance for you to pull constants out of the limit operation. This drastically reduces complexity when dealing with coefficients in your functions. In our problem’s context, the addition of the constant 3 to the product of functions shows this rule as it relates to the sum of limits.
\[\begin{equation}\lim_{{x \to a}} [c \cdot f(x)] = c \cdot \lim_{{x \to a}} f(x)\end{equation}\]
What makes this rule so helpful is its allowance for you to pull constants out of the limit operation. This drastically reduces complexity when dealing with coefficients in your functions. In our problem’s context, the addition of the constant 3 to the product of functions shows this rule as it relates to the sum of limits.
Power of Limit Law
Exponentiation is another common operation in calculus, and handling limits with exponents is guided by the power of limit law. It’s succinctly described as:
\[\begin{equation}\lim_{{x \to a}} [f(x)]^n = [\lim_{{x \to a}} f(x)]^n\end{equation}\]
Essentially, if you're taking the limit of a function raised to a power, you can find the limit of the function first, and then raise that limit to the power. It’s important to note that n should be a real number. This simplifies the calculation process when dealing with functions subjected to powers or roots, as shown in the final step of our example problem. The cube root of the limit is smoothly taken by applying this law, thus finding the limit of the entire expression elegantly.
\[\begin{equation}\lim_{{x \to a}} [f(x)]^n = [\lim_{{x \to a}} f(x)]^n\end{equation}\]
Essentially, if you're taking the limit of a function raised to a power, you can find the limit of the function first, and then raise that limit to the power. It’s important to note that n should be a real number. This simplifies the calculation process when dealing with functions subjected to powers or roots, as shown in the final step of our example problem. The cube root of the limit is smoothly taken by applying this law, thus finding the limit of the entire expression elegantly.
Other exercises in this chapter
Problem 24
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Use the precise definition of a limit to prove the following limits. $$\left.\lim _{x \rightarrow 3}(x-3)^{2}=0 \text { (Hint: Use the identity } \sqrt{x^{2}}=|
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Determining limits analytically Determine the following limits or state that they do not exist. $$\lim _{x \rightarrow 0} \frac{x^{3}-5 x^{2}}{x^{2}}$$
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Determine \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) for the following rational functions. Then give the horizontal asympto
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