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

• May 14th 2009, 02:41 AM
fardeen_gen
[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.
• May 14th 2009, 02:57 AM
josipive
Quote:

Originally Posted by fardeen_gen
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...
• May 14th 2009, 10:11 AM
Moo
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 $\exp\left(x\cdot\frac{f'(x)}{f(x)}\right)$
• May 14th 2009, 12:31 PM
pankaj
The limit is of the type $1^{\infty}$

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

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

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

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

= $exp{ \left\{\frac{xf'(x)}{f(x)}\right\}}$