Find the values of a and b in a limit

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Homework Help Overview

The discussion revolves around finding the values of constants \(a\) and \(b\) in the context of a limit involving a function \(f\). The limit expression is given as \(\lim_{h\rightarrow 0} af(h)+bf(2h)−f(0)=0\) along with the equation \(a+b=1\). Participants are exploring methods to derive a second equation involving \(a\) and \(b\).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using Taylor's theorem to express \(f(h)\) and \(f(2h)\) in terms of \(h\) and explore the implications of this approach. There is also a suggestion to consider the decomposition of the limit into derivatives of \(f\) as an alternative method. Some participants express uncertainty about the applicability of Taylor's theorem based on their coursework.

Discussion Status

The discussion is active, with participants sharing different approaches and hints. Some have attempted to apply Taylor's theorem, leading to a derived equation \(a+2b=0\). Others are considering alternative methods and clarifying definitions related to derivatives, indicating a productive exploration of the problem.

Contextual Notes

Some participants note that their lessons have not covered Taylor's theorem, which may limit their ability to apply this method effectively. There is a focus on understanding the definitions and properties of derivatives as they relate to the limit in question.

songoku
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Homework Statement
Please see below
Relevant Equations
Limit

L'Hopital Rule is not allowed
1664334550562.png


I know $$\lim_{h\rightarrow 0} af(h)+bf(2h)−f(0)=0$$
$$a+b=1$$

But I don't know how to find the second equation involving a and b. I imagine I need to somehow obtain ##h## in numerator so I can cross out with ##h## in denominator but I don't have idea how to get ##h## in the numerator.

Thanks
 
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Try using Taylor's theorem to express ##f(h)## as a linear polynomial in ##h## plus an error term that decreases superlinearly, and do the same for ##f(2h)##.
 
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Office_Shredder said:
Try using Taylor's theorem to express ##f(h)## as a linear polynomial in ##h## plus an error term that decreases superlinearly, and do the same for ##f(2h)##.
Based on the definition of derivative that he gave in another thread, I think this is not the approach that is expected.
 
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FactChecker said:
Based on the definition of derivative that he gave in another thread, I think this is not the approach that is expected.
Yes, my lesson has not covered Taylor's theorem

Office_Shredder said:
Try using Taylor's theorem to express ##f(h)## as a linear polynomial in ##h## plus an error term that decreases superlinearly, and do the same for ##f(2h)##.
I tried using Taylor's theorem, as suggested, as best as I can and by only taking the term up until ##h## I get ##a+2b=0##

Is there other possible method to solve the question without using Taylor?

Thanks
 
songoku said:
Yes, my lesson has not covered Taylor's theoremI tried using Taylor's theorem, as suggested, as best as I can and by only taking the term up until ##h## I get ##a+2b=0##

Is there other possible method to solve the question without using Taylor?

Thanks
Why not do the obvious decomposition of the limit into derivatives of ##f##?
 
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PeroK said:
Why not do the obvious decomposition of the limit into derivatives of ##f##?
I think I understand this hint.

Thank you very much for the help Office_Shredder, FactChecker, PeroK
 
songoku said:
PeroK said:
Why not do the obvious decomposition of the limit into derivatives of ##f##?

I think I understand this hint.
Just in case you don't quite understand that hint, or the advice from @FactChecker in Post #3 ...

You had the following as one of the definitions for ##f'(x_0)##.

##\displaystyle f'(x_0)=\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h} ##

So that ##\displaystyle \ f'(0)=\lim_{h \to 0} \frac{f(h)-f(0)}{h} \ .##

You may not realize that ##\displaystyle \ \lim_{h \to 0} \frac{f(2h)-f(0)}{2h} = f'(0)\ ## as well.

(Added a short time later with the Edit feature):

In decomposing the given limit, take the ##\displaystyle \frac{b\,f(2h)}{h} ## term, for instance and write it as ##\displaystyle \frac{b\,(f(2h)-f(0))}{h} + \frac{b\,(f(0))}{h} \ .##

Do similarly with ##\displaystyle \frac{a\,f(h)}{h}##, combine like terms, etc.
 
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