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Thread: [SOLVED] Prove that limit exists and is non-zero for each x?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Prove that limit exists and is non-zero for each x?

    Let $\displaystyle f$ be a positive differentiable function on $\displaystyle (0,\infty)$

    Prove that $\displaystyle \lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}\right\}^{\frac{1}{h}}$ exists and is non-zero for each x.
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    Quote Originally Posted by fardeen_gen View Post
    Let $\displaystyle f$ be a positive differentiable function on $\displaystyle (0,\infty)$

    Prove that $\displaystyle \lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}\right\}^{\frac{1}{h}}$ exists and is non-zero for each x.
    i will give you a hint...

    $\displaystyle \lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}\right\}^{\frac{1}{h}} = $$\displaystyle \lim_{ h\rightarrow 0} e^{{\ln}{\left\{\frac{f( x + xh)}{f(x)}\right\}}^{\frac{1}{h}}} = $$\displaystyle \lim_{ h\rightarrow 0} e^{{\frac{1}{h}}{\ln}{\left\{\frac{f( x + xh)}{f(x)}\right\}}} = $


    my problem is that i dont know latex...
    Last edited by josipive; May 14th 2009 at 03:24 AM.
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  3. #3
    Moo
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    Then note that $\displaystyle f(x+xh)=f(x)+xh f'(x)+o(h)$

    So $\displaystyle \frac{f(x+xh)}{f(x)}=1+xh \cdot \frac{f'(x)}{f(x)}+o(h)$

    And we know that $\displaystyle \ln(1+x)=x+o(x)$

    Hence $\displaystyle \exp\left(\tfrac 1h \ln\left(1+xh\cdot\tfrac{f'(x)}{f(x)}+o(h)\right)\ right)=\exp\left(x\cdot\tfrac{f'(x)}{f(x)}+o(1)\ri ght)$

    And thus the limit is $\displaystyle \exp\left(x\cdot\frac{f'(x)}{f(x)}\right)$
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  4. #4
    Senior Member pankaj's Avatar
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    The limit is of the type $\displaystyle 1^{\infty}$

    $\displaystyle
    \lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}\right\}^{\frac{1}{h}}$

    $\displaystyle
    =exp(\lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}-1\right\}\frac{1}{h})
    $

    $\displaystyle
    =exp(\lim_{h\rightarrow 0} \left\{\frac{f(x + xh)-f(x)}{hf(x)}\right\})
    $

    $\displaystyle
    =exp(\lim_{h\rightarrow 0} \left\{\frac{xf'(x + xh)}{f(x)}\right\})
    $ (Using L'Hospital's Rule)

    =$\displaystyle exp{ \left\{\frac{xf'(x)}{f(x)}\right\}}$
    Last edited by pankaj; May 14th 2009 at 06:06 PM.
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