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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'Hospital2012-09-17

3 Answers 3

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$ \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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Hint

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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    @kaitlyn No problem at all. What matters is that you understood the concepts. Glad to know this. :)2012-09-18