Extra credit Abstract limit question

[mtayab1994]

Homework Statement

Let f be a derivable function at 0 and f'(0)=2 and let a and b in ℝ.

Calculate the limit: \lim_{x\rightarrow0}\frac{f(ax)-f(bx)}{x}

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.

 

[SammyS]

Homework Statement

Let f be a derivable function at 0 and f'(0)=2 and let a and b in ℝ.

Calculate the limit: \lim_{x\rightarrow0}\frac{f(ax)-f(bx)}{x}

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.

 

[InfinityZero]

The limit is of indeterminate form \frac{0}{0} so you can apply L'Hopital's rule to it to get the limit
\lim_{x\rightarrow0}\frac{af'(ax)-af'(bx)}{1}

which since we know f'(0)=2

should be easy to evaluate.

 

[Dick]

You could also do it directly. Add and subtract f(0) in the numerator, split it up and apply changes of variable like u=ax and v=bx.

 

[SammyS]

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 \frac{0}{0} so you can apply L'Hopital's rule to it to get the limit
\lim_{x\rightarrow0}\frac{af'(ax)-af'(bx)}{1}

which since we know f'(0)=2

should be easy to evaluate.

That's not what I intended.

Use:

\displaystyle f'(0)=\lim_{h\to0}\ \frac{f(0+h)-f(0)}{h}
​

Of course that's the same as \displaystyle f'(0)=\lim_{x\to0}\ \frac{f(x)-f(0)}{x}\ .

Now try using that in the limit you're trying to evaluate. (In the way Dick mentioned.)

 

[mtayab1994]

That's not what I intended.

Use:

\displaystyle f'(0)=\lim_{h\to0}\ \frac{f(0+h)-f(0)}{h}
​

Of course that's the same as \displaystyle f'(0)=\lim_{x\to0}\ \frac{f(x)-f(0)}{x}\ .

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 :).