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I have to find the $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ I know that the result is $f'(a)$, but how do I get until there?

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closed as not a real question by Will Hunting, TMM, rschwieb, Norbert, Cameron Buie Nov 11 '12 at 23:54

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possible duplicate of Limit question mixed with derivatives...? –  Cameron Buie Nov 11 '12 at 23:54
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4 Answers 4

I'm going to assume that your definition of the derivative at $a$ is

$$\lim_{h\rightarrow 0} \frac{f(a+h) - f(a)}{h}.$$

So substituting $x = a + h$ we get

$$ f'(a) = \lim_{h\rightarrow 0} \frac{f(a+h) - f(a)}{h} = \lim_{x - a \rightarrow 0}\frac{f(x) - f(a)}{a + h - a} = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}$$

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This is the definition of the derivative.

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I presume your definition of derivative is $$ f^{'}(a)=\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}. $$

Try to do a change of variable in this to get what you want.

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$$\lim_ {x\to a} \frac{f(x)-f(a)}{x-a}=f'(a)$$ is definition of derivative of function $f$ at $x=a$

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