What is the Limit Definition of a Tough Derivative?

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SUMMARY

The limit definition of a tough derivative is expressed as $$\lim_{h \to 0} \frac{f(x+3h^2) - f(x-h^2)}{2h^2}$$. Participants in the discussion utilized L'Hôpital's Rule to simplify the expression effectively. The hints provided indicate that the function values at the limit approach zero, confirming the derivative's behavior at that point. This approach highlights the importance of understanding limit definitions in calculus.

PREREQUISITES
  • Understanding of calculus concepts, specifically limits and derivatives.
  • Familiarity with L'Hôpital's Rule for evaluating indeterminate forms.
  • Knowledge of function notation and manipulation.
  • Basic algebra skills for handling polynomial expressions.
NEXT STEPS
  • Study the application of L'Hôpital's Rule in various limit scenarios.
  • Explore advanced derivative concepts, such as higher-order derivatives.
  • Learn about Taylor series expansions and their relation to derivatives.
  • Investigate the implications of limit definitions in real-world applications.
USEFUL FOR

Students of calculus, mathematics educators, and anyone looking to deepen their understanding of derivatives and limit definitions.

member 428835
hey pf!

can you help me with this $$\lim_{h \to 0} \frac{f(x+3h^2) - f(x-h^2)}{2h^2}$$

i know the definition and have tried several substitutions, but no help. anyone have any ideas?
 
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nevermind, lopitals rule did the trick
 
hint 1
$$0=\mathrm{f}(x)-\mathrm{f}(x)$$
hint 2
$$\lim_{h \to 0} \frac{f(x+3h^2) - f(x-h^2)}{2h^2}=\lim_{h \to 0}\left[\frac{3}{2}\frac{\mathrm{f}(x+3h^2)-\mathrm{f}(x)}{3h^2}+\frac{1}{2}\frac{\mathrm{f}(x-h^2)-\mathrm{f}(x)}{-h^2}\right]$$
 

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