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Let $f : [0,\infty] \to \mathbb{R}$ be a function such that its derivative $f'(x)=e^{x^2}$.

Compute $$\lim_{x\to 0} \frac{f(2)-f(x+2)}{x}$$

The definition of the derivative is

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

so if $f'(a)=e^{a^2}$ then

$$e^{a^2}=\lim_{x\to 0} \frac{f(a+x)-f(a)}{x}$$

if $a=2$ then

$$e^{a^2}=\lim_{x\to 0} \frac{f(2+x)-f(2)}{x}$$

I'm a little bit stumped here, my naive approach would be to multiply both sides by $(-1)$ and get that the answer is $-e^{2^2}$ but I don't think I can do that.

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    $\begingroup$ What is $-f'(2)$? $\endgroup$ Jul 9, 2018 at 16:22
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    $\begingroup$ @LordSharktheUnknown is it $-e^4$ $\endgroup$ Jul 9, 2018 at 16:23
  • $\begingroup$ $\lim af(x) =a\lim f(x)$ for constant $a$ $\endgroup$
    – kingW3
    Jul 9, 2018 at 16:25

2 Answers 2

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We have $$f'(a)= \lim_{x\to 0}\frac{f(a+x)-f(a)}x$$ but you made a mistake in replacing the LHS with $e^{x^2}$. It should be $e^{a^2}$: $$e^{a^2}=\lim_{x\to 0}\frac{f(a+x)-f(a)}x$$ Set $a=2$ and negate: $$-e^{2^2}=\lim_{x\to0}\frac{f(2)-f(x+2)}x=-e^4$$

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By an elementary property of the limit, and by the definition of the derivative,

$$\lim_{x\to 0} \frac{f(2)-f(x+2)}{x}=-\lim_{x\to 0} \frac{f(x+2)-f(x)}{x}=-f'(2).$$

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