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Given a differentiable function , then what is

$$\lim_{x\to0} \left({ \frac{f(a)}{f(a+x)}}\right)^{\frac2x} $$

Where $a$ is a real number.

If I use the identities $ \ln(1+x) \sim x $

and $ f(a+x) \sim f(a)+xf'(a) $

and take logarithms to both sides my guess is that the limit is

$ \exp\left(- 2\frac{f'(a)}{f(a)}\right) .$

Is this method correct with this result ? thanks in advance

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2 Answers 2

up vote 4 down vote accepted

You can do it that way, with a lot more details.

Alternatively, you can just note that if $g(y)=\log f(y)$, then:

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

But $$e^{\frac{g(x+a)-g(a)}{x}} = \left(\frac{f(x+a)}{f(a)}\right)^{\frac{1}{x}}$$

So the $$\lim_{x\to 0} \left(\frac{f(x+a)}{f(a)}\right)^{\frac{1}{x}} = e^\frac{f'(a)}{f(a)}$$

by continuity of $e^z$. Finally, your limit is this expression raised to the $-2$ power, so your limit is:$$e^{-2\frac{f'(a)}{f(a)}}$$

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(That was probably too much detail for a "homework" answer, sorry.) –  Thomas Andrews Nov 15 '12 at 18:57

The answer is correct. You can make your method rigorous using big O notation.

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