Question

Help me solve this

Answer 1

Prove that if f(x) has derivative at x=a, then \[\lim_{x \rightarrow 0} (xf(a)-af(x))/(x-a) = f(a) - af'(x)\]

Answer 2

You mean x->a, right?

Answer 3

no x --> 0

Answer 4

Are you absolutely sure? Because I just solved it for x->a. Ill write down my solution anyway for people to work with if possible.

Answer 5

Nevermind, I did not :P Ill fix it, but you should be right

Answer 6

yes, I absolutely sure. But you post your answer please

Answer 7

\[just \ \ notice \ \ \\xf(a)-af(x)=xf(a)-af(x)+af(a)-af(a)=(x-a)f(a)-a(f(x)-f(a))\]

Answer 8

\[f(a) - a f'(x) = f(a) - a \lim_{h \rightarrow 0}{f(x+h)-f(x) \over h} =\]
\[f(a) - \lim_{h \rightarrow 0}{af(x+h)-af(x) \over h} = \lim_{h \rightarrow 0}{ hf(a) - af(x+h)+af(x) \over h}\]
Make h = a-x
So, if h->0, x->a, making it 0 = a-a = 0. Thus
\[\lim_{h \rightarrow 0}{ hf(a) - af(x+h)+af(x) \over h} = \lim_{x \rightarrow a}{ (a-x)f(a) - af(x+(a-x))+af(x) \over a-x}=\]
\[\lim_{x \rightarrow a}{ af(a)-xf(a) - af(a)+af(x) \over a-x} = \lim_{x \rightarrow a}{ -xf(a)+af(x) \over a-x} \]
Multiply both sides by -1
\[ \lim_{x \rightarrow a}{ xf(a)-af(x) \over x-a} \]

Answer 9

Woops. I did it wrong again. X->a. my bad. lol, thats my solution anyway, I dont get it where Im doing it wrong

Answer 10

Oh, I think this has to do with f'(x) being diferentiable at x=a. You probably have to use that..

Answer 11

Thank anyway @soati :)

Answer 12

I think I got it. If f'(a) exists then does that means f'(a) = constant?

Answer 13

\[\lim_{x \rightarrow 0} \frac{xf(a)-af(x)}{x-a}=\lim_{x \rightarrow 0} \frac{xf(a)-af(x)+af(a)-af(a)}{x-a}\\=\lim_{x \rightarrow 0} \frac{(x-a)f(a)-a(f(x)-f(a))}{x-a}=\lim_{x \rightarrow 0} f(a)-a\frac{f(x)-f(a)}{x-a}=f(a)-a\lim_{x \rightarrow 0} \frac{f(x)-f(a)}{x-a}\\ is \ \ \lim_{x \rightarrow 0} \frac{f(x)-f(a)}{x-a} \ \ equal \ \ to \ \ f'(x) \ \ ??!!! \]

Answer 14

No, only if it was x-> a :(

Answer 15

if it was x-> a the limit will be f'(a) !!!!

Answer 16

so sorry all i got alittle confuse
\[limx→0(xf(a)−af(x))/(x−a)=f(a)−af′(a)\]

Answer 17

^^"

Answer 18

could u plz check the question again
i think it must be lim x->a

Answer 19

Well, if you take my solution and replace x by a inside the first limit then you s hould get the solution, no?

Answer 20

@mukushla : i only confuse with f'(x) with f'(a) , but lim still x-> 0

Answer 21

ok

Answer 22

@soati : i replace x by a inside the first limit but still x -> a not x -> 0 :(

Answer 23

I had to leave to do something. Anyway, I got this:

\[\lim_{x \rightarrow 0}{xf(a) - af(x) \over x-a} = f(a)-af´(a)\]
\[\lim_{x \rightarrow 0}{xf(a) - af(x) \over x-a} = \lim_{x \rightarrow 0}{xf(a) +af(a) -af(a)- af(x) \over x-a}=\lim_{x \rightarrow 0}{(x-a)f(a) + a(f(a)-f(x)) \over x-a} =\]
\[\lim_{x \rightarrow 0}{(x-a)f(a) + a(f(a)-f(x)) \over x-a} = \lim_{x \rightarrow 0}f(a) + {a(f(a)-f(x)) \over x-a} =f(a) - a \lim_{x \rightarrow 0}{f(x)-f(a) \over x-a}\]
\[f(a) - a {f(0)-f(a) \over -a} = f(a) +f(0) -f(a) = f(0)\]

However if you use Eulers method, which states that \[f(x+h) = f(x) + hf´(x)\]
Which is just an approximation for nonlinear functions... However if we assume that our f is linear, then eulers equation holds. This is why I earlier asked if f'(a) existing could mean it was a constant. Because that would mean f(x) is linear.

And so from Eulers method \[f(a-a) = f(0) = f(a) - af´(a)\]
Since we got f(0) as our solution to the limit, we get that the first statement is true IF f(x) is linear.

I dont know how to prove it otherwise. I think that this proof working means f(x) really is linear, otherwise eulers method would just give us an approximation, not the exact answer as in this case.

Answer 24

@soati : Thank you so much :)