Problem 73
Question
Continuous Functions In Exercises 73 and \(74,\) find the value of \(c\) that makes the function continuous at \(x=0\). \(f(x)=\left\\{\begin{array}{ll}\frac{4 x-2 \sin 2 x}{2 x^{3}}, & x \neq 0 \\\ c, & x=0\end{array}\right.\)
Step-by-Step Solution
Verified Answer
The constant \(c\) that makes the function continuous at \(x = 0\) is \(c = 2\).
1Step 1: Calculate the Limit of \(f(x)\) as \(x\) Approaches Zero
The first step is to calculate \(\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{4x - 2\sin{2x}}{2x^3}\). Using L'Hopital's rule, which states that the limit of a ratio of two functions is equal to the limit of the ratios of their derivatives, we get \(\lim_{{x \to 0}} \frac{4 - 4\cos{2x}}{6x^2}\). Again applying L'Hopital's rule, this becomes \(\lim_{{x \to 0}} \frac{8\sin{2x}}{12x}\), which simplifies further to \(\lim_{{x \to 0}} \frac{\sin{2x}}{x} = 2\) using the standard limit \(\lim_{{x \to 0}} \frac{\sin{x}}{x} = 1\).
2Step 2: Equate the Limit of \(f(x)\) as \(x\) Approaches Zero to \(f(0)\)
To ensure the function is continuous at \(x = 0\), its value at that point must match the limit as \(x\) approaches zero. From step 1, we know \(\lim_{{x \to 0}} f(x) = 2\). The function is defined to be equal to \(c\) at \(x = 0\), therefore, to make the function continuous at \(x = 0\), \(c\) must be equal to 2.
3Step 3: Conclusion
After analyzing the limit of the function as \(x\) approaches zero and the function value at \(x = 0\), it is concluded that the function is continuous at \(x = 0\) when \(c = 2\). This is the value that makes the function continuous at \(x = 0\).
Key Concepts
L'Hopital's RuleLimit of a FunctionContinuity at a Point
L'Hopital's Rule
When we encounter limits that result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hopital's Rule becomes a valuable tool. It helps in evaluating these limits by simplifying the limits through differentiation.
The rule states that for functions \(f(x)\) and \(g(x)\) that are differentiable in an interval around \(c\) (except possibly at \(c\) itself), if \(\lim_{{x \to c}} f(x) = \lim_{{x \to c}} g(x) = 0\) or both are infinite, then:
By taking the derivatives of the numerator and the denominator step by step, with L'Hopital's Rule applied twice, the limit was simplified. This approach yielded a concrete limit value, helping confirm the value for \(c\) to ensure continuity at \(x = 0\).
The rule states that for functions \(f(x)\) and \(g(x)\) that are differentiable in an interval around \(c\) (except possibly at \(c\) itself), if \(\lim_{{x \to c}} f(x) = \lim_{{x \to c}} g(x) = 0\) or both are infinite, then:
- \(\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}\) provided the limit on the right side exists.
By taking the derivatives of the numerator and the denominator step by step, with L'Hopital's Rule applied twice, the limit was simplified. This approach yielded a concrete limit value, helping confirm the value for \(c\) to ensure continuity at \(x = 0\).
Limit of a Function
The limit of a function is a fundamental concept in calculus that describes the behavior of a function as it approaches a specific point. When we say \(\lim_{{x \to a}} f(x) = L\), it means that as \(x\) gets closer to \(a\), \(f(x)\) gets closer to \(L\).
Understanding limits helps in analyzing and predicting the behavior of functions at points where they might not be explicitly defined.
In our exercise, we focused on evaluating \(\lim_{{x \to 0}} \frac{4x - 2\sin{2x}}{2x^3}\), which helped in determining the value of \(c\) for continuity.
By evaluating this limit using L'Hopital's Rule, we were able to conclude that the limit as \(x\) approaches zero is 2. This provided the critical information needed for selecting \(c\) such that the function becomes seamless and continuous at \(x = 0\).
Understanding limits helps in analyzing and predicting the behavior of functions at points where they might not be explicitly defined.
In our exercise, we focused on evaluating \(\lim_{{x \to 0}} \frac{4x - 2\sin{2x}}{2x^3}\), which helped in determining the value of \(c\) for continuity.
By evaluating this limit using L'Hopital's Rule, we were able to conclude that the limit as \(x\) approaches zero is 2. This provided the critical information needed for selecting \(c\) such that the function becomes seamless and continuous at \(x = 0\).
Continuity at a Point
Continuity at a point means that a function behaves predictably without any jumps, breaks, or holes at that point. For a function \(f(x)\) to be continuous at a point \(x = a\), three conditions must be satisfied:
We first ensured the limit of \(f(x)\) as \(x\) approached zero was computed. This limit was discovered to be 2.
Next, by setting \(f(0) = c = 2\), the continuity conditions were satisfied, ensuring \(f(x)\) smoothly passed through \(x = 0\). This approach showed the nice interaction between limit calculation and the definition of continuity.
- \(f(a)\) is defined.
- \(\lim_{{x \to a}} f(x)\) exists.
- \(f(a) = \lim_{{x \to a}} f(x)\).
We first ensured the limit of \(f(x)\) as \(x\) approached zero was computed. This limit was discovered to be 2.
Next, by setting \(f(0) = c = 2\), the continuity conditions were satisfied, ensuring \(f(x)\) smoothly passed through \(x = 0\). This approach showed the nice interaction between limit calculation and the definition of continuity.
Other exercises in this chapter
Problem 72
Find the centroid of the region bounded by the graphs of \(f(x)=x^{2}, g(x)=2^{x}, x=2,\) and \(x=4\)
View solution Problem 73
Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f\) is continuous on \([0, \infty)
View solution Problem 73
Use integration by parts to verify the reduction formula. $$ \int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x $$
View solution Problem 73
Integrals Used to Find Fourier Coefficients \(\quad\) In Exercises 73 and \(74,\) verify the value of the definite integral, where \(n\) is a positive integer.
View solution