The problem involves calculating the limit of a difference quotient for a derivable function at zero, specifically \(\lim_{x\rightarrow0}\frac{f(ax)-f(bx)}{x}\), given that the derivative at zero is \(f'(0)=2\). The context is rooted in calculus, focusing on limits and derivatives.
The discussion is active, with participants offering different methods and questioning the correctness of initial answers. Some participants suggest re-evaluating the limit using the definition of the derivative, while others explore the implications of the indeterminate form.
There is a noted confusion regarding the correct form of the answer, with multiple interpretations being considered. The original poster expresses uncertainty about their initial solution, prompting further exploration of the problem.
2a-b is not the correct answer. 2(a-b), or a-b may possibly be the answer.mtayab1994 said:Homework Statement
Let f be a derivable function at 0 and f'(0)=2 and let a and b in ℝ.
Calculate the limit: [tex]\lim_{x\rightarrow0}\frac{f(ax)-f(bx)}{x}[/tex]
The Attempt at a Solution
I'm not sure but i got 2a-b as my answer but i want to know how to solve it the proper way any help is very much appreciated.
SammyS said:2a-b is not the correct answer. 2(a-b), or a-b may possibly be the answer.
Write f ' (0) as a limit & see where you need to go from there.
That's not what I intended.InfinityZero said:The limit is of indeterminate form [tex]\frac{0}{0}[/tex] so you can apply L'Hopital's rule to it to get the limit
[tex]\lim_{x\rightarrow0}\frac{af'(ax)-af'(bx)}{1}[/tex]
which since we know [tex]f'(0)=2[/tex]
should be easy to evaluate.
SammyS said:That's not what I intended.
Use:[itex]\displaystyle f'(0)=\lim_{h\to0}\ \frac{f(0+h)-f(0)}{h}[/itex]Of course that's the same as [itex]\displaystyle f'(0)=\lim_{x\to0}\ \frac{f(x)-f(0)}{x}\ .[/itex]
Now try using that in the limit you're trying to evaluate. (In the way Dick mentioned.)