Definition of the Derivative using delta and epsilon

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SUMMARY

The discussion clarifies the definition of the derivative using the epsilon-delta formulation. It confirms that if the limit exists, one can express it as: for all ε > 0, there exists a δ > 0 such that |x - x₀| < δ implies |(f(x) - f(x₀))/(x - x₀) - f'(x₀)| < ε. This equivalence reinforces the foundational concepts of calculus and the rigorous approach to limits.

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  • Understanding of limits in calculus
  • Familiarity with the epsilon-delta definition of limits
  • Basic knowledge of derivatives
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jfy4
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Hi,

I have a question about the formulation of the derivative. The definition is

[tex]f'(x_0)=\lim_{x\rightarrow x_0}\frac{f(x)-f(x_0)}{x-x_0}[/itex]<br /> <br /> Lets say this limit exists. Can I write the limit in the typical [itex]\epsilon-\delta[/itex] method as such<br /> <br /> <i>Given the limit exists, then for all [itex]\epsilon>0[/itex] there exists a [itex]\delta>0[/itex] such that [itex]|x-x_0|<\delta\implies[/itex]</i><br /> [tex]\left|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)\right|<\epsilon[/tex]<br /> <br /> ?[/tex]
 
Physics news on Phys.org
Yes, the word "limit" and the "lim" notation mean just the same here as they usually do.
 
Rasalhague said:
Yes, the word "limit" and the "lim" notation mean just the same here as they usually do.

Thanks
 

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