Problem 4
Question
If \(f(x)=\frac{x-e^{x}+\cos 2 x}{x^{2}}, x \neq 0\) is continuous at \(x=0\), then (a) \(f(0)=\frac{5}{2}\) (b) \([f(0)]=-2\) (c) \(\\{f(0)\\}=-0.5\) (d) \([f(0)] \cdot\\{f(0)\\}=-1.5\) (where \([x]\) and \(\\{x\\}\) denotes greatest integer and fractional part function.)
Step-by-Step Solution
Verified Answer
(b), (c), and (d) are true.
1Step 1: Evaluate Limit
Since \(f\) is continuous at \(x=0\), the limit of \(f(x)\) as \(x\) approaches \(0\) must equal \(f(0)\). Therefore, we need to find the limit: \[\lim_{{x \to 0}} \frac{x - e^{x} + \cos 2x}{x^2}.\]
2Step 2: Use L'Hôpital's Rule
The limit is of the form \(\frac{0}{0}\), so we can use L'Hôpital's Rule. We differentiate the numerator and the denominator:Derivative of the numerator: \(1 - e^x - 2\sin 2x\).Derivative of the denominator: \(2x\).Now apply L'Hôpital's Rule:\[\lim_{{x \to 0}} \frac{1 - e^x - 2\sin 2x}{2x}.\]
3Step 3: Apply L'Hôpital's Rule Again
The limit \(\frac{1 - e^x - 2\sin 2x}{2x}\) is still of the form \(\frac{0}{0}\), so apply L'Hôpital's Rule again. Differentiate the new numerator and denominator:Derivative of the new numerator: \(-e^x - 4\cos 2x\).Derivative of the denominator: 2.Evaluate the new limit:\[\lim_{{x \to 0}} \frac{-e^x - 4\cos 2x}{2}.\]
4Step 4: Evaluate the Final Limit
Substitute \(x=0\) into the differentiated expression:\[\frac{-e^0 - 4\cos 0}{2} = \frac{-1 - 4}{2} = \frac{-5}{2}.\]Since \(f\) is continuous at \(x=0\), we have \(f(0) = \frac{-5}{2}\).
5Step 5: Verify Options
Now check the given options with \(f(0) = \frac{-5}{2}\).(a) False: The value is \(-\frac{5}{2}\), not \(\frac{5}{2}\).(b) True: The greatest integer less than or equal to \(-\frac{5}{2}\) is \(-3\).(c) True: The fractional part of \(-\frac{5}{2}\) is \(-0.5\), since \(-\frac{5}{2} = -3 + 0.5\).(d) True: \([f(0)] \cdot \{f(0)\} = (-3) \cdot (-0.5) = 1.5.\)
Key Concepts
Limit EvaluationL'Hôpital's RuleGreatest Integer FunctionFractional Part Function
Limit Evaluation
When discussing continuity in calculus, one of the core principles is the evaluation of limits. Limits help us understand the behavior of a function as the input approaches a specific value. In the given exercise, we're interested in evaluating \ \( \lim_{{x \to 0}} \frac{x - e^{x} + \cos 2x}{x^2} \ \). This step is crucial because if a function is continuous at a point, the limit of the function as it approaches that point must equal the function's value at that point.
Using limit evaluation, we focus on understanding what happens to the function \ \( f(x) \ \) as \ \( x \ \) gets arbitrarily close to zero.
To begin, plug in values of \ \( x \ \) that get nearer to zero. You might notice the expression initially seems to approach a \ \( \frac{0}{0} \ \) indeterminacy form. This is where techniques, like L'Hôpital's Rule, become useful to refine our approach.
Using limit evaluation, we focus on understanding what happens to the function \ \( f(x) \ \) as \ \( x \ \) gets arbitrarily close to zero.
To begin, plug in values of \ \( x \ \) that get nearer to zero. You might notice the expression initially seems to approach a \ \( \frac{0}{0} \ \) indeterminacy form. This is where techniques, like L'Hôpital's Rule, become useful to refine our approach.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that result in indeterminate forms, such as \ \( \frac{0}{0} \ \) or \ \( \frac{\infty}{\infty} \ \). The rule states that if the limits of both the numerator and the denominator of a fraction are zero or infinity, the limit of the fraction is the same as the limit of the derivatives of the numerator and the denominator.
In the exercise, after evaluating the initial limit, we realized it was of the form \ \( \frac{0}{0} \ \), which allows us to apply L'Hôpital's Rule.
To use the rule, differentiate the numerator, \ \( 1 - e^x - 2\sin 2x \ \), and the denominator, \ \( 2x \ \), and take the limit again.
When this new limit still results in \ \( \frac{0}{0} \ \), L'Hôpital's Rule can be applied once more, leading us to the final expression \ \( \frac{-e^x - 4\cos 2x}{2} \ \), resulting in a computable limit. Thus, L'Hôpital's Rule helps bridge us from an indeterminate form to a clear solution.
In the exercise, after evaluating the initial limit, we realized it was of the form \ \( \frac{0}{0} \ \), which allows us to apply L'Hôpital's Rule.
To use the rule, differentiate the numerator, \ \( 1 - e^x - 2\sin 2x \ \), and the denominator, \ \( 2x \ \), and take the limit again.
When this new limit still results in \ \( \frac{0}{0} \ \), L'Hôpital's Rule can be applied once more, leading us to the final expression \ \( \frac{-e^x - 4\cos 2x}{2} \ \), resulting in a computable limit. Thus, L'Hôpital's Rule helps bridge us from an indeterminate form to a clear solution.
Greatest Integer Function
The greatest integer function, denoted as \ \([x]\ \), is the function that takes a real number and gives you the largest integer less than or equal to that number. It's also known as the floor function.
For example, \ \([3.7] = 3\ \), \ \( [-2.1] = -3 \ \).
When determining \ \([f(0)] \ \) for the given problem, where \ \( f(0) = \frac{-5}{2} \ \), which is \ \(-2.5\ \), we find that \ \([f(0)] \ \) becomes \ \(-3\ \) because it is the largest integer that is less than or equal to \ \(-2.5\ \).
This concept is crucial in determining step-by-step solutions where approximation and precision around integers play a role in problem-solving.
For example, \ \([3.7] = 3\ \), \ \( [-2.1] = -3 \ \).
When determining \ \([f(0)] \ \) for the given problem, where \ \( f(0) = \frac{-5}{2} \ \), which is \ \(-2.5\ \), we find that \ \([f(0)] \ \) becomes \ \(-3\ \) because it is the largest integer that is less than or equal to \ \(-2.5\ \).
This concept is crucial in determining step-by-step solutions where approximation and precision around integers play a role in problem-solving.
Fractional Part Function
The fractional part function, expressed as \ \(\{x\}\ \), focuses on the decimal or fractional component of a real number. It is defined as the non-integer part of a number \ \( x \ \), essentially \ \( x - [x] \ \).
With these functions, it's easier to breakdown and understand expressions involving non-integer quantities, like the options presented in the exercise, which require calculating parts of expressions beyond simple integers.
- For \ \( 3.7 \ \), the fractional part is \ \( \{3.7\} = 0.7\ \).
- For \ \( -2.5 \ \), using the greatest integer function, we find \ \(-2.5 = -3 + 0.5 \ \).
With these functions, it's easier to breakdown and understand expressions involving non-integer quantities, like the options presented in the exercise, which require calculating parts of expressions beyond simple integers.
Other exercises in this chapter
Problem 1
For \(x>0\), let \(h(x)=\left\\{\begin{array}{ll}\frac{1}{q}, & \text { if } x=\frac{p}{q} \\ 0, & \text { if } x \text { is irrational }\end{array}\right.\) wh
View solution Problem 7
Let \([x]\) denotes the integral part of \(x \in R \cdot g(x)=x-[x]\). Let \(f(x)\) be any continuous function with \(f(0)=f(1)\), then the function \(h(x)=f(g(
View solution Problem 8
Let \(f\) be a differentiable function on the open interval \((a, b)\), Which of the following statements must be true? I. \(f\) is continuous on the closed int
View solution Problem 10
). Consider function \(f: R-\\{-1,1\\} \rightarrow R . f(x)=\frac{x}{1-|x|}\). Then, the incorrect statement is (a) it is continuous at the origin (b) it is not
View solution