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I know that $f'(x)$ exists at $x = 1$, and moreover that $f(1) = K > 0$.

I need to calculate the limit

$$\lim_{x\rightarrow1} \left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}$$

How could I calculate this ?

Thanks in advance !

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de l'Hospital might be a way to go... –  Fabian Jan 7 '13 at 19:26
    
hi @Fabian , how could l'Hospital help in this case ? I can't see it, maybe you can clarify it for me –  itamar Jan 7 '13 at 19:28
    
I provided an answer... –  Fabian Jan 7 '13 at 19:33

3 Answers 3

up vote 5 down vote accepted

Expand the function around $x=1$ using the derivative and the series expansion of $\log (x+1)$

$$\lim_{\Delta x\rightarrow 0} \left(\frac{f(1)+\Delta x f'(1)}{f(1)}\right) ^ {1/log(1+\Delta x)} = \lim_{\Delta x\rightarrow 0} \left(1+\Delta x\frac{f'(1)}{f(1)}\right) ^ {1/\Delta x}$$ $$=e^{f'(1)/f(1)}$$

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You can bring it into the form $0/0$ via $$ \lim_{x\to1}\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)} = \exp \left(\lim_{x\to1} \frac{\log [f(x)/f(1)]}{\log x} \right)$$

De l'Hospital then gives $$\lim_{x\to1}\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)} = \exp \left( \lim_{x\to1}\frac{x f'(x)}{f(x)} \right)$$

The limit can then be evaluated and we have $$\lim_{x\to1}\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)} =\exp\left(\frac{f'(1)}{f(1)}\right).$$

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Let $$y=\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}$$

So, $$\log y=\frac{\log f(x)-\log f(1)}{\log x}=\frac{\log f(x)-\log f(1)}{f(x)-f(1)}\frac{f(x)-f(1)}{x-1}\frac{x-1}{\log x-\log 1}$$

$$\lim_{x\to1}\log y=\left(\frac{d\log f(x)}{d f(x)}\frac{d f(x)}{dx}\frac1{\frac{d\log x}{dx}}\right)_{x=1}=\left(\frac1{f(x)}\cdot{f'(x)}\cdot x\right)_{x=1}=\frac{f'(1)}{f(1)}$$

So, $$\lim_{x\to1} \left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}=\lim_{x\to1} y=e^{\frac{f'(1)}{f(1)}}$$

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