Problem 64 - Calculus Single Variable

Question

Prove that the function $$ f(x)=\left\\{\begin{array}{lll} x^{2} \cdot \sin (1 / x) & \text { if } & x \neq 0 \\ 0 & \text { if } & x=0 \end{array}\right. $$ is differentiable at \(x=0,\) and compute \(f^{\prime}(0)\).

Step-by-Step Solution

Answer

The function is differentiable at \( x = 0 \), and \( f'(0) = 0. \)

Step 1: Understand Differentiability at a Point

To prove that function \( f(x) \) is differentiable at \( x = 0 \), we need to check if the derivative \( f'(0) \) exists. The derivative at \( x = 0 \) is given by the definition: \[ f'(0) = \lim_{{h \to 0}} \frac{{f(h) - f(0)}}{h}. \] Since \( f(0) = 0 \), the expression simplifies to \[ f'(0) = \lim_{{h \to 0}} \frac{{f(h)}}{h}. \]

Step 2: Substitute the Function into Derivative Expression

For \( x eq 0 \), \( f(x) = x^2 \sin(1/x) \). Substitute this into the expression for the derivative: \[ f'(0) = \lim_{{h \to 0}} \frac{{h^2 \sin(1/h)}}{h} = \lim_{{h \to 0}} h \sin(1/h). \]

Step 3: Evaluate the Limit

Now evaluate the limit \( \lim_{{h \to 0}} h \sin(1/h) \). The key observation here is that \( \sin(1/h) \) is bounded by -1 and 1. Therefore, \( -h \leq h \sin(1/h) \leq h \). As \( h \to 0 \), both \( -h \) and \( h \) approach zero. Applying the squeeze theorem gives: \[ \lim_{{h \to 0}} h \sin(1/h) = 0. \]

Step 4: Conclude Differentiability and Compute the Derivative

Since \( \lim_{{h \to 0}} h \sin(1/h) = 0 \), the derivative \( f'(0) \) exists and equals 0. Thus, the function \( f(x) \) is differentiable at \( x = 0 \) with \( f'(0) = 0. \)

Key Concepts

Squeeze Theorem

The squeeze theorem is a valuable tool in calculus for evaluating limits that are otherwise difficult to handle. This theorem helps when you can establish that a function is squeezed between two other functions whose limits at a certain point are known. To understand this better, consider three functions: \( g(x) \), \( f(x) \), and \( h(x) \). Suppose you know that \( g(x) \leq f(x) \leq h(x) \) for all \( x \) close to but not necessarily at a given point, and both \( \lim_{x \to a} g(x) = L \) and \( \lim_{x \to a} h(x) = L \). Then, the squeeze theorem tells us that \( \lim_{x \to a} f(x) = L \) as well.

In the given problem, we used the squeeze theorem to find \( \lim_{h \to 0} h \sin(1/h) \). Since the sine function is bounded by -1 and 1, we know that \( -h \leq h \sin(1/h) \leq h \). Since both \( -h \) and \( h \) converge to 0 as \( h \to 0 \), the squeeze theorem allows us to conclude that \( \lim_{h \to 0} h \sin(1/h) = 0 \). The insight here is that by using functions that bound the problematic function, you can deduce the behavior of the limit.

• Important for simplifying limit problems.
• Relies on bounding a function by two others with the same limit at a point.

Limit Evaluation

Evaluating limits is a fundamental skill in calculus, involving the behavior of a function as the input approaches a particular value. Limits are crucial for defining derivatives and integrals. In limit evaluation, different strategies, such as direct substitution, factoring, rationalization, or the squeeze theorem, are used based on the function’s complexity and form.

In the exercise, we wanted to find \( \lim_{h \to 0} h \sin(1/h) \) to determine the derivative of \( f(x) \) at \( x = 0 \). By observing that \( \sin(1/h) \) only takes values between -1 and 1, it becomes manageable to bound the expression and make the evaluation feasible via the squeeze theorem.

• Identifying the correct technique is crucial based on the limit form.
• Recognizing bounded functions assists in applying different strategies.

Limit evaluation not only helps in derivative computation but also in identifying function continuity and asymptotic behavior.

Derivative at a Point

A derivative represents the rate at which a function is changing at any given point. Mathematically, it is the slope of the tangent line to the curve at that point. To find the derivative of a function at a specific point \( x = a \), we use the definition, which is \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \). Through this process, we assess the function’s behavior right around \( x = a \).

In the problem statement, proving that \( f(x) \) is differentiable at \( x = 0 \) involved showing that \( f'(0) \) exists. This was done by finding the limit \( \lim_{h \to 0} h \sin(1/h) \), which turned out to be 0. Hence, \( f'(0) \) exists and equals 0.

• The derivative of a function at a point tells us the instantaneous rate of change.
• It is vital in understanding the function’s behavior near the specific point.

Understanding derivatives at a point is essential in analyzing how the function behaves and changes as values approach this point.