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

1. ## [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.

2. Originally Posted by fardeen_gen
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...

3. 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 $\displaystyle \exp\left(x\cdot\frac{f'(x)}{f(x)}\right)$

4. 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)

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