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Math Help - [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 f be a positive differentiable function on (0,\infty)

    Prove that \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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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Let f be a positive differentiable function on (0,\infty)

    Prove that \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...

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

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

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

    Hence \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 \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 1^{\infty}

     <br />
\lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}\right\}^{\frac{1}{h}}

     <br />
=exp(\lim_{h\rightarrow 0} \left\{\frac{f(x + xh)}{f(x)}-1\right\}\frac{1}{h})<br />

     <br />
=exp(\lim_{h\rightarrow 0} \left\{\frac{f(x + xh)-f(x)}{hf(x)}\right\})<br />

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

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