Help Solve Calculus Limit Proof Homework Statement

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SUMMARY

The forum discussion focuses on proving the limit of a function as it approaches zero, specifically addressing the statement "f(x) is differentiable at x=0." The user suggests rewriting the function using the expression f(x) = f(0) + x (f'(0) + \epsilon(x)), where \epsilon(x) approaches zero as x approaches zero. This approach clarifies the behavior of the function near the limit and provides a structured way to analyze the proof. The thread also notes that it is a duplicate of a previous discussion on the same topic.

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  • Understanding of calculus concepts, specifically limits and derivatives.
  • Familiarity with the definition of differentiability at a point.
  • Knowledge of epsilon-delta definitions in limit proofs.
  • Basic algebraic manipulation skills for rewriting functions.
NEXT STEPS
  • Study the epsilon-delta definition of limits in calculus.
  • Learn how to apply the concept of differentiability in limit proofs.
  • Explore examples of limit proofs involving derivatives, particularly at points of discontinuity.
  • Review advanced calculus techniques for handling indeterminate forms.
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Students studying calculus, particularly those tackling limit proofs and differentiability, as well as educators looking for examples to illustrate these concepts.

JasMath33
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Homework Statement


I am posting this for another student who I noticed did not have the proof in the problem. Here is what she said. Let's try and help her out.

I have been working on the problem below and I am stuck. I am stuck primarily because of the part where is says x=0. If x-0, it should cancel everything out. The derivative of 0 is 0 so will cancel everything out I think, so I am not sure if that is the reasoning and the proof behind it.

cap-2-png.102533.png


Homework Equations


She had no relevant equations.

The Attempt at a Solution


She said:

I thought I should start this way, but I am not 100% sure.
upload_2016-6-26_20-51-50-png.102534.png

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The meaning of "f(x) is differentiable at x=0" is that near x=0, we can write:

\dfrac{f(x) - f(0)}{x} = f'(0) + \epsilon(x)

where \epsilon(x) is a correction term that goes to zero as x \rightarrow 0. So we can rewrite f(x) as:

f(x) = f(0) + x (f'(0) + \epsilon(x))

So try using that expression to rewrite f(ax) and f(bx).
 

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