Proving Derivative of sin(x)/x is Zero at x=0

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Discussion Overview

The discussion revolves around proving that the derivative of the function sin(x)/x is zero at x=0. The scope includes mathematical reasoning and the application of calculus concepts such as limits and differentiability.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the definition of differentiability and applying l'Hopital's rule or series expansion to prove the limit exists.
  • Another participant points out that f(0) = sin(0)/0 is not defined, raising a concern about the initial function's definition at that point.
  • A later reply proposes that the discussion may be about the sinc function, which extends the definition of f(x) = sin(x)/x to include f(0) = 1 by continuity.

Areas of Agreement / Disagreement

Participants express differing views on the definition of the function at x=0, with some suggesting the use of the sinc function while others focus on the original function sin(x)/x. The discussion remains unresolved regarding the approach to proving the derivative.

Contextual Notes

There are limitations regarding the assumptions about the function's definition at x=0 and the applicability of l'Hopital's rule or series expansion without resolving the undefined nature of sin(0)/0.

mherna48
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Hey can anyone help me prove that the derivative of sin(x)/x is zero at x=0
 
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A function f:R -> R is differentiable at 0 if
\lim_{h \rightarrow 0}\frac{f(h)-f(0)}{h}
exists. You can apply this definition and then use l'Hopital's rule or perhaps some first-order estimates to show that this last limit exists. Alternatively, you can probably work directly with the series expansion for sin.
 
snipez90 said:
A function f:R -> R is differentiable at 0 if
\lim_{h \rightarrow 0}\frac{f(h)-f(0)}{h}
exists. You can apply this definition and then use l'Hopital's rule or perhaps some first-order estimates to show that this last limit exists. Alternatively, you can probably work directly with the series expansion for sin.
`
But f(0) = sin0/0 is not defined
 
Well I took it that the OP actually meant the sinc function, which just extends f(x) = sin(x)/x by continuity so that f(0) = 1.
 

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