Extra credit Abstract limit question

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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 wan't to know how to solve it the proper way any help is very much appreciated.
 
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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.
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.
 
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:
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.

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.
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.)
 
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.)

Wow thanks a lot i didn't realize it was that simple :).