Find the values of a and b in a limit

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The discussion revolves around finding the values of a and b in the limit equation involving a function f. The initial equation provided is $$\lim_{h\rightarrow 0} af(h)+bf(2h)−f(0)=0$$ along with the constraint $$a+b=1$$. Participants suggest using Taylor's theorem to express f(h) and f(2h) in terms of h, leading to the derived equation $$a+2b=0$$. Some contributors propose an alternative approach by decomposing the limit into derivatives of f, which may simplify the problem without relying on Taylor's theorem. The conversation emphasizes the importance of understanding the definitions of derivatives in solving the limit equation.
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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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