Math Problem Statement
Determine whether f'(0) exists for the following functions:
59. f(x) = {x * sin(1/x) if x ≠ 0; 0 if x = 0}
60. f(x) = {x^2 * sin(1/x) if x ≠ 0; 0 if x = 0}
Solution
To determine if exists for both functions in problems 59 and 60, we need to examine the differentiability at . We will analyze the definition of the derivative at , which is:
Since in both cases, this simplifies to:
Let's evaluate this limit for each function individually.
Problem 59
The function is given by:
x \sin \left( \frac{1}{x} \right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$$ To find $$f'(0)$$, we need to compute: $$f'(0) = \lim_{x \to 0} \frac{x \sin \left( \frac{1}{x} \right)}{x} = \lim_{x \to 0} \sin \left( \frac{1}{x} \right)$$ The function $$\sin \left( \frac{1}{x} \right)$$ oscillates between -1 and 1 as $$x \to 0$$, meaning the limit does not settle to a single value. Thus, the limit does not exist, and $$f'(0)$$ does not exist for this function. ### Problem 60 The function is given by: $$f(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$$ To find $$f'(0)$$, we compute: $$f'(0) = \lim_{x \to 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{x} = \lim_{x \to 0} x \sin \left( \frac{1}{x} \right)$$ Now, $$x \sin \left( \frac{1}{x} \right)$$ is bounded because $$\sin \left( \frac{1}{x} \right)$$ oscillates between -1 and 1, so: $$- |x| \leq x \sin \left( \frac{1}{x} \right) \leq |x|$$ As $$x \to 0$$, both $$-|x|$$ and $$|x|$$ approach 0. By the Squeeze Theorem: $$\lim_{x \to 0} x \sin \left( \frac{1}{x} \right) = 0$$ Therefore, $$f'(0) = 0$$ for this function, so $$f'(0)$$ exists and is equal to 0. ### Summary 1. For problem 59, $$f'(0)$$ does not exist. 2. For problem 60, $$f'(0) = 0$$. Would you like more details or have any questions on these calculations? Here are 5 related questions to consider: 1. How does oscillation affect the existence of a limit? 2. What other conditions can make a limit exist or not? 3. Can you determine the continuity of these functions at $$x = 0$$? 4. How would $$x^n \sin \left( \frac{1}{x} \right)$$ behave for different values of $$n$$? 5. What if we replace $$\sin \left( \frac{1}{x} \right)$$ with $$\cos \left( \frac{1}{x} \right)$$ — would $$f'(0)$$ still exist? **Tip:** The Squeeze Theorem is particularly useful for analyzing limits of functions that oscillate but are bounded by functions approaching zero.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Differentiability
Limits
Oscillating Functions
Squeeze Theorem
Formulas
Definition of derivative: f'(0) = lim(x -> 0) [f(x) - f(0)] / x
Squeeze Theorem: if g(x) <= f(x) <= h(x) and lim(x -> a) g(x) = lim(x -> a) h(x) = L, then lim(x -> a) f(x) = L
Theorems
Squeeze Theorem
Suitable Grade Level
Undergraduate Calculus