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Hi guys,

Starting revision for my exams and I can't seem to solve these problems. ):

1. Given that three non zero vectors

2. In another question, given that f(x) is differenciatable at x = 0 and [itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a, wjere a and k are constants.

b. Show that f'(0) = a/1-k

What I did was take the defination of differenciation:

[itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

and I let x = kx, and h = (1-k)x,

[itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

and I did this (assuming that I can write [itex]\frac{Lim}{(1-k)x->0}[/itex] as [itex]\frac{Lim}{x->0}[/itex]

1/(1-k)[itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a/1-k

Is this correct?

Thanks in advance for your help guys!

*Edit* Just realized that I misread the forum title and posted in the wrong one. ):

Starting revision for my exams and I can't seem to solve these problems. ):

1. Given that three non zero vectors

**A**,**B**and**C**are perpendicular to each other and that the vector**R**= p**A**+ q**B**+ r**C**is a zero vector, show that p = q = r = 02. In another question, given that f(x) is differenciatable at x = 0 and [itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a, wjere a and k are constants.

b. Show that f'(0) = a/1-k

What I did was take the defination of differenciation:

[itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

and I let x = kx, and h = (1-k)x,

[itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

and I did this (assuming that I can write [itex]\frac{Lim}{(1-k)x->0}[/itex] as [itex]\frac{Lim}{x->0}[/itex]

1/(1-k)[itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a/1-k

Is this correct?

Thanks in advance for your help guys!

*Edit* Just realized that I misread the forum title and posted in the wrong one. ):

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