Limit

**chaddy** · February 12th 2008, 08:49 AM

If f is differentiable everywhere, then, in terms of f and a, what is the value of: lim x-->a (f(f(x))-f(f(a)))/(x-a)?

**ecMathGeek** · February 12th 2008, 10:37 AM

That equation simply finds the slope of a secant line between the points (x,f(x)) to (a,f(a)) as x approaches a. When x approaches a, the slope of the secant line begins to approach the slop of f(x) at a. In other words, it finds the slope at a or the derivative at a. Therefore, it equals f'(a).

**mr fantastic** · February 12th 2008, 02:19 PM

Originally Posted by chaddy

If f is differentiable everywhere, then, in terms of f and a, what is the value of: lim x-->a (f(f(x))-f(f(a)))/(x-a)?

Originally Posted by ecMathGeek

That equation simply finds the slope of a secant line between the points (x,f(x)) to (a,f(a)) as x approaches a. When x approaches a, the slope of the secant line begins to approach the slop of f(x) at a. In other words, it finds the slope at a or the derivative at a. Therefore, it equals f'(a).

I'm afraid not. Unless the f(f(x)) and f(f(a)) are typos, the answer is the derivative of f(f(x)) at x = a, NOT the derivative of f(x) at x = a ......

Using the chain rule:

$[f(f(x))]' = f'(f(x)) \times f'(x)$ . At x = a, this will equal $f'(f(a)) \times f'(a)$ . Making it a little neater, if you knew that f(a) = b, then it will equal $f'(b) \times f'(a)$ ......

**chaddy** · February 12th 2008, 02:28 PM

Thanks for catching that.

**Plato** · February 12th 2008, 03:35 PM

This question is a bit more complicated, I think.
Because f is differentiable everywhere it is continuous everywhere.
That means that $\lim _{x \to a} \left( {f(x) - f(a)} \right) = 0$ .
Now consider the following:
$\lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}<br /> {{x - a}} = \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}<br /> {{f(x) - f(a)}}\lim _{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}<br /> {{x - a}}$
$\lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}<br /> {{x - a}} = f'\left( {f(a)} \right)f'(a)$

**ecMathGeek** · February 12th 2008, 03:45 PM

I looked at it too quickly to notice the double f's. Thanks for catching that, Mr. Fantastic.

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