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To evaluate this type of limits, how can I do, considering $f$ differentiable, and $ f (x_0)> 0 $

$$\lim_{x\to x_0} \biggl(\frac{f(x)}{f(x_0)}\biggr)^{\frac{1}{\ln x -\ln x_0 }},\quad\quad x_0>0,$$

$$\lim_{x\to x_0} \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0},\quad\quad n\in\mathbb{N}.$$

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4 Answers 4

up vote 2 down vote accepted

For the first: for $x\neq x_0$, since $f(x)>0$ in a neighborhood of $x_0$ \begin{align*} \frac{f(x)}{f(x_0)}^{\frac 1{\ln x-\ln x_0}}&=\exp\left(\frac {\ln f(x)-\ln f(x_0)}{\ln x-\ln x_0}\right)\\ &=\exp\left(\frac {\ln f(x)-\ln f(x_0)}{x-x_0}\frac{x-x_0}{\ln x-\ln x_0}\right), \end{align*} and taking the limit $x\to x_0$, since $\ln f$ is differentiable, we get $$\lim_{x\to x_0}\frac{f(x)}{f(x_0)}^{\frac 1{\ln x-\ln x_0}}=\exp\left(\frac{f'(x_0)}{f(x_0)}\frac 1{\frac 1{x_0}}\right)=\exp\left(x_0\frac{f'(x_0)}{f(x_0)}\right).$$ For the second question \begin{align*}\frac{x_0^nf(x)-x^nf(x_0)}{x-x_0}&=x_0^n\frac{f(x)-f(x_0)}{x-x_0}+f(x_0)\frac{x_0^n-x^n}{x-x_0}\\ &=x_0^n\frac{f(x)-f(x_0)}{x-x_0}-f(x_0)\sum_{k=0}^{n-1}x^kx_0^{n-k-1}, \end{align*} and taking the limit $x\to x_0$ we get $$\lim_{x\to x_0}\frac{x_0^nf(x)-x^nf(x_0)}{x-x_0}=x_0^nf'(x_0)-nf(x_0)x_0^{n-1}.$$

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Hint: write $x=x_0+h$ with $h\to0$ and expand each term up to order $h$.

For example, $\log(x)-\log(x_0)=\log(1+h/x_0)=h/x_0$ and $f(x)=f(x_0)+hf'(x_0)$ hence $f(x)/f(x_0)=1+hf'(x_0)/f(x_0)$, hence...

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For the second limit you can also observe that if $x_0 \neq 0$ then

$$\lim_{x\to x_0} \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0} = \lim_{x \to x_0}x^nx_0^n \frac{ \frac{f(x)}{x^n}- \frac{f(x_0)}{x_0^n}}{x-x_0}= (x_0)^{2n} (\frac{f(x)}{x^n})'(x_0) \,.$$

The case $x_0=0$ is trivial.

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For the first problem, one may be tempted to use L'Hopital's Rule:

Noting that $f(x)>0$ for $x$ sufficiently close to $x_0$:

$$\ln \biggl[\,\Bigl ({f(x)\over f(x_0)}\Bigr)^{1\over \ln x-\ln x_0}\,\biggr] = {\ln f(x)-\ln f(x_0)\over \ln x-\ln x_0}.$$

We have: $$\tag{1}\lim_{x\rightarrow x_0} {\ln f(x)-\ln f(x_0)\over \ln x-\ln x_0} = \lim_{x\rightarrow x_0} {{f'(x)/ f(x)} \over 1/x} =\lim_{x\rightarrow x_0} { {xf'(x)\over f(x)} }={ {x_0f'(x_0)\over f(x_0)} }. $$

So $$\lim_{x\rightarrow x_0}\Bigl ({f(x)\over f(x_0)}\Bigr)^{1\over \ln x-\ln x_0}= \exp\Bigl({ {x_0 f'(x_0)\over f(x_0)} }\Bigr). $$

But, this line of reasoning is incorrect. We do not know that the last limit appearing in (1) exists.

See the other answers for correct arguments.

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Would this be better as a comment? –  David Mitra Dec 31 '11 at 1:20
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