Problem 22
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. \(\lim _{x \rightarrow 1} \frac{f(x)}{g(x)-h(x)}\)
Step-by-Step Solution
Verified Answer
Answer: The limit of the given function as x approaches 1 is 8.
1Step 1: Identify the function in the given limit
The given limit is
$$
\lim_{x \rightarrow 1} \frac{f(x)}{g(x) - h(x)}
$$
2Step 2: Identify the limits of the numerator and denominator
The numerator is f(x) and the denominator is g(x) - h(x). The limits of these functions as x approaches 1 are given in the problem:
$$
\lim_{x \rightarrow 1} f(x) = 8 \\
\lim_{x \rightarrow 1} g(x) = 3 \\
\lim_{x \rightarrow 1} h(x) = 2
$$
3Step 3: Apply the sum/difference limit law to denominator
Using the difference limit law, which states that the limit of a difference is the difference of the limits, we can compute the limit of the denominator:
$$
\lim_{x \rightarrow 1} (g(x) - h(x)) = \lim_{x \rightarrow 1} g(x) - \lim_{x \rightarrow 1} h(x) = 3 - 2 = 1 \\
$$
4Step 4: Apply the quotient limit law
Applying the quotient limit law, which states that the limit of a quotient is the quotient of the limits, provided that the limit of the denominator is nonzero, we get:
$$
\lim_{x \rightarrow 1} \frac{f(x)}{g(x) - h(x)} = \frac{\lim_{x \rightarrow 1} f(x)}{\lim_{x \rightarrow 1}(g(x) - h(x))} = \frac{8}{1} = 8 \\
$$
The limit of the given function as x approaches 1 is 8, and we used the difference and quotient limit laws to compute it.
Key Concepts
Quotient Limit LawDifference Limit LawLimits of Functions
Quotient Limit Law
Understanding the quotient limit law is essential when dealing with limits involving division. In calculus, the quotient limit law states that if the limit of two functions exists as x approaches a value, then the limit of their quotient also exists and is equal to the quotient of their limits. This is formally written as:
\[\begin{equation}\lim_{{x \to a}} \frac{{f(x)}}{{g(x)}} = \frac{{\lim_{{x \to a}} f(x)}}{{\lim_{{x \to a}} g(x)}}\end{equation}\]
where \[\begin{equation}\lim_{{x \to a}} g(x) eq 0.\end{equation}\]
This law is instrumental in simplifying complex expressions by allowing us to divide the limits of individual functions rather than dealing with the limit of a fraction as a whole. Bear in mind, however, that this rule only applies when the denominator's limit isn't zero, as division by zero is undefined. In the given exercise, we applied this rule to evaluate the limit of \[\begin{equation}\frac{{f(x)}}{{g(x) - h(x)}}\end{equation}\]as x approaches 1, ensuring a smooth and simplified approach to solving the problem.
\[\begin{equation}\lim_{{x \to a}} \frac{{f(x)}}{{g(x)}} = \frac{{\lim_{{x \to a}} f(x)}}{{\lim_{{x \to a}} g(x)}}\end{equation}\]
where \[\begin{equation}\lim_{{x \to a}} g(x) eq 0.\end{equation}\]
This law is instrumental in simplifying complex expressions by allowing us to divide the limits of individual functions rather than dealing with the limit of a fraction as a whole. Bear in mind, however, that this rule only applies when the denominator's limit isn't zero, as division by zero is undefined. In the given exercise, we applied this rule to evaluate the limit of \[\begin{equation}\frac{{f(x)}}{{g(x) - h(x)}}\end{equation}\]as x approaches 1, ensuring a smooth and simplified approach to solving the problem.
Difference Limit Law
When approaching limits involving subtraction, the difference limit law becomes a valuable tool for calculation. This law is straightforward and asserts that the limit of the difference between two functions is the difference of their limits. Expressed mathematically:
\[\begin{equation}\lim_{{x \to a}} (f(x) - g(x)) = \lim_{{x \to a}} f(x) - \lim_{{x \to a}} g(x).\end{equation}\]
This allows us to dissect complex expressions into more manageable pieces. For instance, in our example, we applied the difference limit law to find the limit of the denominator \[\begin{equation}(g(x) - h(x))\end{equation}\]as x approaches 1. This simplification made it possible to easily compute each function's limit separately and then subtract them to find the overall limit of the denominator.
\[\begin{equation}\lim_{{x \to a}} (f(x) - g(x)) = \lim_{{x \to a}} f(x) - \lim_{{x \to a}} g(x).\end{equation}\]
This allows us to dissect complex expressions into more manageable pieces. For instance, in our example, we applied the difference limit law to find the limit of the denominator \[\begin{equation}(g(x) - h(x))\end{equation}\]as x approaches 1. This simplification made it possible to easily compute each function's limit separately and then subtract them to find the overall limit of the denominator.
Limits of Functions
The core concept of limits in calculus is the behavior of a function as the input (often denoted as x) approaches a certain value. Limits help us understand and deal with situations involving continuity and change, and form the foundation for derivative calculations. The fundamental idea is captured by the notation:\[\begin{equation}\lim_{{x \to a}} f(x),\end{equation}\]
which represents the value that the function f(x) approaches as x gets infinitely close to 'a'. Limits are not always equal to the function's value at that point, which allows us to deal with functions that are not well-defined at certain points.
In practice, calculating limits often involves sum, difference, product, quotient, and other limit laws. These laws enable the rearrangement and breaking down of complex expressions into more manageable parts, each of which can be approached individually. In our example, by using limit laws correctly, we were able to calculate the limits of f(x), g(x), and h(x) independently and then apply these laws to compute the final limit of the complex expression \[\begin{equation}\frac{{f(x)}}{{g(x) - h(x)}}\end{equation}\]as x approaches 1. This process demonstrated not only the power of limit laws but also how they play a critical role in simplifying the learning and application of calculus.
which represents the value that the function f(x) approaches as x gets infinitely close to 'a'. Limits are not always equal to the function's value at that point, which allows us to deal with functions that are not well-defined at certain points.
Application to Calculations
In practice, calculating limits often involves sum, difference, product, quotient, and other limit laws. These laws enable the rearrangement and breaking down of complex expressions into more manageable parts, each of which can be approached individually. In our example, by using limit laws correctly, we were able to calculate the limits of f(x), g(x), and h(x) independently and then apply these laws to compute the final limit of the complex expression \[\begin{equation}\frac{{f(x)}}{{g(x) - h(x)}}\end{equation}\]as x approaches 1. This process demonstrated not only the power of limit laws but also how they play a critical role in simplifying the learning and application of calculus.
Other exercises in this chapter
Problem 22
Determine the following limits. $$\lim _{x \rightarrow-\infty} 2 x^{-8}$$
View solution Problem 22
Use the precise definition of a limit to prove the following limits. $$\lim _{x \rightarrow 3} \frac{x^{2}-7 x+12}{x-3}=-1$$
View solution Problem 22
Use Theorem 2.10 to determine the intervals on which the following functions are continuous. $$g(x)=\frac{3 x^{2}-6 x+7}{x^{2}+x+1}$$
View solution Problem 23
Determining limits analytically Determine the following limits or state that they do not exist. a. \(\lim _{x \rightarrow 2^{+}} \frac{x^{2}-4 x+3}{(x-2)^{2}}\)
View solution