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$$\lim_{x \to 5} \frac{f(x^2)-f(25)}{x-5}$$

Assuming that $f$ is differentiable for all $x$, simplify.

(It does not say what $f(x)$ is at all)

My teacher has not taught us any of this, and I am unclear about how to proceed.

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Hint: l'Hospital – celtschk Sep 17 '12 at 21:15
up vote 2 down vote accepted

$$ \frac{f(x^2)-f(25)}{x-5} = \frac{f(x^2)-f(25)}{x^2-25} \cdot (x+5)$$

Since, $f$ is differentiable, if $x\to 5$ then $x^2\to 25$, so taking the lim will give you $f'(25)\cdot 10$.

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Apply Lagrange theorem to $f(x^2)$

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$f$ is differentiable, so $g(x) = f(x^2)$ is also differentiable. Let's find the derivative of $g$ at $x = 5$ using the definition.

$$ g'(5) = \lim_{x \to 5} \frac{g(x) - g(5)}{x - 5} = \lim_{x \to 5} \frac{f(x^2) - f(25)}{x - 5} $$

Now write $g'(5)$ in terms of $f$ to get the desired result.

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We haven't even learned about derivatives :( – kaitlyn Sep 17 '12 at 21:23
@kaitlyn - how then can you interpret a question which says that $f$ is differentiable as a key piece of information - what do you understand by that? – Mark Bennet Sep 17 '12 at 21:27
that is whole issue here, I am trying to teach myself this information right now, by teaching myself derivatives, and what I understand so far is that the derivative of f exists, and that it implies continuity – kaitlyn Sep 17 '12 at 21:43
@kaitlyn I think you need to be a little more comfortable with differentiation than you are to attempt this kind of question... Do you at least know the definition of a derivative? i.e. $f'(a)=\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. Can you see how this relates to your question? Once this is clear the answers on this page should make sense. – user39572 Sep 17 '12 at 22:20
@julien yes, I understand that now, I've been studying derivatives and I understand this definition, but now I am just confused about how to write g prime of 5 in terms of f, I understand why he used g(x) to put the limit into the form of the definition of a derivative – kaitlyn Sep 17 '12 at 22:40

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