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Chapter 2: Problem 38

Explain with words and a graph why \(f(x)=x \sin (1 / x)\) is continuous but has no derivative at \(x=0 .\) Set \(f(0)=0 .\)

Short Answer

Expert verified
The function is continuous at \(x=0\) by the Squeeze Theorem but not differentiable as the limit for the derivative oscillates.

Step by step solution

01

Define and analyze the function

The function is given as \(f(x)=x \sin (1 / x)\) for \(x eq 0\), and \(f(0)=0\). We need to analyze its behavior around \(x=0\).
02

Check continuity at x=0

For \(f(x)\) to be continuous at \(x=0\), the limit \(\lim_{x \to 0} f(x) = f(0)\) must hold. Compute \(\lim_{x \to 0} x \sin(1/x)\). Since \(-1 \leq \sin(1/x) \leq 1\), multiplying by \(x\) gives \(-x \leq x \sin(1/x) \leq x\). As \(x\to 0\), both \(-x\) and \(x\) tend towards 0, thus by the Squeeze Theorem, \(\lim_{x \to 0} x \sin(1/x) = 0\), which equals \(f(0)\). Hence, \(f(x)\) is continuous at \(x=0\).
03

Attempt to differentiate at x=0

To check differentiability at \(x=0\), compute the derivative \(f'(x)\) using the definition: \(f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h \sin(1/h)}{h}\). Simplifying gives \(\lim_{h \to 0} \sin(1/h)\), but \(\sin(1/h)\) oscillates between -1 and 1 as \(h\to 0\). Consequently, this limit does not exist, indicating \(f(x)\) is not differentiable at \(x=0\).
04

Visualize using a graph

Plot \(f(x) = x \sin(1/x)\). As \(x\) approaches 0, notice the oscillations shrink in amplitude but remain frequent, highlighting continuity but not differentiability at 0. The graph remains visually connected and approaches a flat line at \(x=0\), suggesting continuity but without a derivative slope, as it doesn't settle to a single value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Squeeze Theorem
Intuitively, the Squeeze Theorem helps us determine the limit of a function that is "squeezed" between two known functions. This is useful when the function itself might behave unpredictably, especially near a point of interest like zero.When we have a function, say, \(f(x) = x \sin(1/x)\), and we want to understand its behavior as \(x\) approaches 0, the Squeeze Theorem can be very handy.We know:
  • \(-1 \leq \sin(1/x) \leq 1\)
  • Multiplying all parts by \(x\), which is approaching zero, gives \(-x \leq x \sin(1/x) \leq x\)
As \(x\) gets closer to 0, both \(-x\) and \(x\) converge to 0. Therefore, by the Squeeze Theorem:\[\lim_{x \to 0} x \sin(1/x) = 0\]This conclusion helps us affirm that the function is continuous at \(x = 0\) because the limit exists and matches \(f(0)\).
Oscillation
Oscillation in this context refers to the behavior of \(\sin(1/x)\) as \(x\) approaches zero. Unlike a predictable gradual approach towards a single value, oscillation implies rapid swings.In the function \(f(x) = x \sin(1/x)\), as \(x\) becomes very small, the value of \(1/x\) becomes very large. This results in \(\sin(1/x)\) oscillating wildly between -1 and 1:
  • High frequency: The oscillations increase in frequency as \(x\) approaches 0.
  • Consistency in extremes: Despite high frequency, \(\sin(1/x)\) remains bounded by -1 and 1.
While the amplitude of the oscillations in \(f(x)\) shrinks because of the multiplier \(x\), this oscillation prevents the derivative at 0 from being defined. The rapid changes mean there's no consistent direction, essential for a derivative.
Discontinuity
Discontinuity occurs when a function doesn't remain unbroken at a point or over an interval. This might involve sharp changes, undefined points, or jumps.In the function \(f(x) = x \sin(1/x)\), although \(f(x)\) is defined for all \(x\) including 0, it might seem tricky without careful assessment. However, despite the wild oscillations near \(x = 0\):
  • The function does not exhibit an actual discontinuity at 0.
  • The Squeeze Theorem proves the continuity by showing \(\lim_{x \to 0} f(x) = f(0)\).
The rapid changes don't break the function鈥檚 continuity at 0. Continuity is retained as the limit behavior echoes the function鈥檚 value at that point.
Limit Behavior
Limit behavior describes how a function behaves as its input approaches a particular point. It鈥檚 crucial in understanding continuity and differentiability.With \(f(x) = x \sin(1/x)\), studying the limit as \(x\) approaches 0 is key:
  • The calculated limit \(\lim_{x \to 0} f(x) = 0\) aligns with \(f(0) = 0\), ensuring continuity.
  • However, differentiability requires more than the function鈥檚 limit aligning with its value at a point.
The behavior of \(\sin(1/x)\) as \(x\to 0\), which oscillates without settling on a stable slope, indicates that \(f(x)\) lacks a clear direction, precluding a defined derivative.Understanding limit behavior through detailed analysis thus reveals not just continuity, but also the nuances of differentiability.

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What is \(a_{4}\) and what is the limit \(L\) ? After which \(N\) is \(\left|a_{n}-L\right|<\frac{1}{10} ?\) (Calculator allowed) (a) \(-1,+\frac{1}{2},-\frac{1}{3}, \ldots\) (b) \(\frac{1}{2}, \frac{1}{2}+\frac{1}{4}, \frac{1}{2}+\frac{1}{4}+\frac{1}{6}, \ldots\) (c) \(\frac{1}{2}, \frac{2}{4}, \frac{3}{8}, \ldots a_{n}=n / 2^{n}\) (d) \(1.1,1.11,1.111, \ldots\) (e) \(a_{n}=\sqrt[n]{n}\) (f) \(a_{n}=\sqrt{n^{2}+n}-n\) (g) \(1+1,\left(1+\frac{1}{2}\right)^{2},\left(1+\frac{1}{3}\right)^{3}, \ldots\)

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