Prove differentiability and continuity

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SUMMARY

The function f(x) is defined as f(x) = {xsin(1/x) if x ≠ 0, 0 if x = 0}. It has been established that f'(0) does not exist due to the oscillatory nature of sin(1/x) as x approaches 0, which confirms that f is not differentiable at x = 0. Furthermore, the discussion concludes that since f'(0) does not exist, f'(x) is not continuous across the real numbers.

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  • Familiarity with derivatives and differentiability
  • Basic trigonometric functions and their properties
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  • Learn about the formal definition of continuity and differentiability
  • Explore the behavior of oscillatory functions near limits
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Students studying calculus, particularly those focusing on differentiability and continuity, as well as educators seeking to explain the nuances of piecewise functions and their derivatives.

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Homework Statement



Determine that, if f(x) =

{xsin(1/x) if x =/= 0
{0 if x = 0

that f'(0) exists and f'(x) is continuous on the reals. (Sorry I can't type the function better, it's piecewise)

Homework Equations



The Attempt at a Solution



For f'(0) existing,

For x ≠ 0,

(f(x) - f(0))/(x - 0) = (x sin(1/x))/x = sin(1/x).

Sin(1/x) doesn't have a limit as x → 0 because the function oscillates between -1 and 1, f is not differentiable at x = 0. Therefore f'(0) doesn't exist.

For f'(x) being continuous.

For any given ɛ > 0, there exists δ > 0 such that |x - y| < δ and |f(x) - f(y)| < ɛ for all x, y on the reals.

For all x [tex]\in[/tex] R (R is the reals), ... and this where I'm stuck.
 
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If you know that [itex]f'(0)[/itex] doesn't exist, doesn't that prove that [itex]f'(x)[/itex] is not continuous on the reals?
 
Yeah, you're right. Haha, I'm sorry. Total brain letdown on that one.
 

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