Problem 32
Question
If \(f\) is differentiable function satisfying \(f(0)=0\), and if \(g(x)=\frac{f(x)}{x}\), then the value, that should be assigned to \(g(0)\), so that \(g\) is continuous at ' 0 ' is (A) 1 (B) 0 (C) \(f(0)\) (D) \(f^{\prime}(0)\)
Step-by-Step Solution
Verified Answer
(D) \(f'(0)\)
1Step 1: Understanding the problem
We are given a differentiable function \( f \) that satisfies \( f(0) = 0 \). We need to find the value to assign to \( g(0) \) for function \( g(x) = \frac{f(x)}{x} \) to be continuous at \( x=0 \). Continuous here means that \( \lim_{{x \to 0}} g(x) = g(0) \).
2Step 2: Expressing the limit as x approaches 0
To ensure that \( g(x) \) is continuous at \( x = 0 \), calculate \( \lim_{{x \to 0}} g(x) \). The expression for \( g(x) \) becomes \( \lim_{{x \to 0}} \frac{f(x)}{x} \). Since \( f \) is differentiable at 0, we apply L'Hopital's rule which requires us to differentiate \( f(x) \) and \( x \).
3Step 3: Applying L'Hopital's Rule
L'Hopital's rule states that for \( \lim_{{x \to 0}} \frac{f(x)}{x} \), we differentiate numerator and denominator to obtain \( \lim_{{x \to 0}} \frac{f'(x)}{1} = f'(0) \). Thus, for the limit to exist and \( g(x) \) to be continuous at 0, we assign \( g(0) \) the value: \( f'(0) \).
4Step 4: Selecting the correct option
The solution states that \( \lim_{{x \to 0}} g(x) = f'(0) \), so the value assigned to \( g(0) \) should be \( f'(0) \). Therefore, the correct option is D.
Key Concepts
Differentiable FunctionsL'Hopital's RuleLimit of a Function
Differentiable Functions
When we describe a function as differentiable, we imply that it is smooth and has a well-defined derivative at every point within its domain. Differentiability is a step up from continuity, meaning that if a function is differentiable at a point, it is also continuous there. However, the reverse is not always true — a continuous function may not be differentiable.
A differentiable function doesn't just have a derivative; the derivative itself varies smoothly. This smooth behavior implies that there's no sharp turns or cusps on the graph of the function.
At the heart of differentiability is the derivative function, denoted as \( f'(x) \), which represents the rate of change of \( f \) at any point. If \( f \) is differentiable at \( x = 0 \), we have \( f'(0) \), which is critical in problems assessing limits and continuity.
A differentiable function doesn't just have a derivative; the derivative itself varies smoothly. This smooth behavior implies that there's no sharp turns or cusps on the graph of the function.
At the heart of differentiability is the derivative function, denoted as \( f'(x) \), which represents the rate of change of \( f \) at any point. If \( f \) is differentiable at \( x = 0 \), we have \( f'(0) \), which is critical in problems assessing limits and continuity.
L'Hopital's Rule
L'Hopital's Rule is a powerful tool used to evaluate limits of indeterminate forms such as \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). This rule becomes particularly useful when a limit results in one of these forms and allows us to differentiate the numerator and the denominator separately to find the limit of their division.
The key is that both functions must be differentiable around the point of interest. Importantly, L'Hopital's Rule assumes that the original problem still leads to an indeterminate form even after initial simplifications.
In applying L'Hopital's Rule to find \( \lim_{{x \to 0}} \frac{f(x)}{x} \), we differentiate \( f(x) \) to get \( f'(x) \) and \( x \) to obtain \( 1 \). Thus, L'Hopital's Rule helps us simplify the limit to \( f'(0) \), indicating the smoothness of \( f \) extends to the behavior of \( \frac{f(x)}{x} \).
The key is that both functions must be differentiable around the point of interest. Importantly, L'Hopital's Rule assumes that the original problem still leads to an indeterminate form even after initial simplifications.
In applying L'Hopital's Rule to find \( \lim_{{x \to 0}} \frac{f(x)}{x} \), we differentiate \( f(x) \) to get \( f'(x) \) and \( x \) to obtain \( 1 \). Thus, L'Hopital's Rule helps us simplify the limit to \( f'(0) \), indicating the smoothness of \( f \) extends to the behavior of \( \frac{f(x)}{x} \).
Limit of a Function
The concept of a limit is fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. Limits help us understand the behavior of functions at points where they may not be explicitly defined.
In the problem context, we are investigating the limit of \( g(x) = \frac{f(x)}{x} \) as \( x \to 0 \). For \( g(x) \) to be continuous at \( x = 0 \), this limit must equal \( g(0) \).
In practical terms, getting \( \lim_{{x \to 0}} \frac{f(x)}{x} = f'(0) \) ensures continuity by providing a means to "fill in" the function where division by zero would otherwise make it undefined. Obtaining this result confirms \( g(0) = f'(0) \), a critical link between the limit, differentiation, and defining behavior smoothly at points near \( x = 0 \).
In the problem context, we are investigating the limit of \( g(x) = \frac{f(x)}{x} \) as \( x \to 0 \). For \( g(x) \) to be continuous at \( x = 0 \), this limit must equal \( g(0) \).
In practical terms, getting \( \lim_{{x \to 0}} \frac{f(x)}{x} = f'(0) \) ensures continuity by providing a means to "fill in" the function where division by zero would otherwise make it undefined. Obtaining this result confirms \( g(0) = f'(0) \), a critical link between the limit, differentiation, and defining behavior smoothly at points near \( x = 0 \).
- Limits examine the behavior of a function near a particular point, rather than at that point.
- Continuity at a point involves the agreement of the value of a function and its limit.
- In the context of \( \lim_{{x \to 0}} g(x) \), ensuring this limit exists and matches \( g(0) \) is key for continuity.
Other exercises in this chapter
Problem 30
The value of \(f(0)\) so that the function \(f(x)=\frac{\sqrt[3]{1+x}-\sqrt[4]{1+x}}{x}\) becomes continuous at \(x=0\), is (A) \(\frac{1}{12}\) (B) \(\frac{7}{
View solution Problem 31
If \(f\) is an even function such that \(\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\) has some finite non-zero value, then (A) \(f\) is continuous and derivabl
View solution Problem 33
Let \(f(x)=\frac{1}{[\sin x]},[.]\) being the greatest integer function, then (A) \(f(x)\) is not continuous, where \(x \in(2 n \pi, 2 n \pi+\pi), n \in I\) (B)
View solution Problem 35
If \(f(x)=x+\frac{x}{1+x}+\frac{x}{(1+x)^{2}}+\ldots\) to \(\infty\), then at \(x=0\), \(f(x)\) (A) has no limit (B) is discontinuous (C) is continuous but not
View solution