find a function from the limit

**jill** · July 20th 2010, 06:40 AM

Suppose

lim_{x -> infty} { log f(x) / log x } = c ,

where c is a constant.

Can we prove that:

f(x) = b x^c + lower order terms,

where b is something independent of x ?

can anyone show me how to prove this? or maybe it is incorrect?

tks

**Failure** · July 20th 2010, 08:12 AM

Originally Posted by jill

Suppose

lim_{x -> infty} { log f(x) / log x } = c ,

where c is a constant.

Can we prove that:

f(x) = b x^c + lower order terms,

where b is something independent of x ?

It doesn't come out like this for me, because if I use Landau's little-oh notation the given limit means that

$\frac{\log f(x)}{\log x}=c+o(1)$ for $x\rightarrow\infty$

in other words

$\log f(x)=\big(c+o(1)\big)\cdot \log x$

and thus

$f(x)=x^{c+o(1)}=x^{o(1)}\cdot x^c$ for $x\rightarrow \infty$

Now, if $o(1)$ goes to 0 veeeery slooowly, $x^{o(1)}$ will actually get arbitrarily large, so: no, b cannot generally be independent of x.

**jill** · July 20th 2010, 09:36 AM

Originally Posted by Failure

It doesn't come out like this for me, because if I use Landau's little-oh notation the given limit means that

$\frac{\log f(x)}{\log x}=c+o(1)$ for $x\rightarrow\infty$

in other words

$\log f(x)=\big(c+o(1)\big)\cdot \log x$

and thus

$f(x)=x^{c+o(1)}=x^{o(1)}\cdot x^c$ for $x\rightarrow \infty$

Now, if $o(1)$ goes to 0 veeeery slooowly, $x^{o(1)}$ will actually get arbitrarily large, so: no, b cannot generally be independent of x.

Thanks a lot!

So that if we know

$\lim_{x \rightarrow \infty} \frac{log f_1(x)}{ log x } = c_1$ , and

$\lim_{x \rightarrow \infty} \frac{log f_2(x)}{ log x } = c_2$ ,

then we can prove (if $c_2>c_1$ ):

$\lim_{x \rightarrow \infty} \frac{ log [ a1 f_1(x) + a2 f_2(x) ] }{ [ log x ] } = c_2 .$

Since

$f_1(x) = x^{ c_1 + o(1) }$ , and

$f_2(x) = x^{ c_2 + o(1) }$ .

so

$\lim_{x \rightarrow \infty} \frac{ log [ a_1 f_1(x) + a_2 f_2(x) ] }{ [ log x ] } =$
$\lim_{x \rightarrow \infty} \frac{ log [ a_1 x^{ c_1 + o(1) } + a_2 x^{ c_2 + o(1) } ] }{ [ log x ] } =$
$\lim_{x \rightarrow \infty} \frac{ log [ a_2 x^{ c_2 + o(1) } ( \frac{a_1}{a_2} x^{ c_1 -c_2 + o(1) } + 1 ) ] }{ [ log x ] } =$
$\lim_{x \rightarrow \infty} \frac{ log a_2 + log [x^{ c_2 + o(1) }] + log 1 }{ [ log x ] } =$
$c_2$

please correct me if anything is wrong

**Failure** · July 20th 2010, 10:37 AM

Originally Posted by jill

Thanks a lot!

So that if we know

$\lim_{x \rightarrow \infty} \frac{log f_1(x)}{ log x } = c_1$ , and

$\lim_{x \rightarrow \infty} \frac{log f_2(x)}{ log x } = c_2$ ,

then we can prove (if $c_2>c_1$ ):

$\lim_{x \rightarrow \infty} \frac{ log [ a1 f_1(x) + a2 f_2(x) ] }{ [ log x ] } = c_2 .$

Since

$f_1(x) = x^{ c_1 + o(1) }$ , and

$f_2(x) = x^{ c_2 + o(1) }$ .

so

$\lim_{x \rightarrow \infty} \frac{ log [ a_1 f_1(x) + a_2 f_2(x) ] }{ [ log x ] } =$
$\lim_{x \rightarrow \infty} \frac{ log [ a_1 x^{ c_1 + o(1) } + a_2 x^{ c_2 + o(1) } ] }{ [ log x ] } =$
$\lim_{x \rightarrow \infty} \frac{ log [ a_2 x^{ c_2 + o(1) } ( \frac{a_1}{a_2} x^{ c_1 -c_2 + o(1) } + 1 ) ] }{ [ log x ] } =$
$\lim_{x \rightarrow \infty} \frac{ log a_2 + log [x^{ c_2 + o(1) }] + log 1 }{ [ log x ] } =$
$c_2$

please correct me if anything is wrong

Looks ok to me. - But I didn't quite figure that this is what you really wanted to prove.

**jill** · July 20th 2010, 10:48 AM

Originally Posted by Failure

Looks ok to me. - But I didn't quite figure that this is what you really wanted to prove.

This really is what i want to prove.
It looks must-be-true, but i did get stuck when i tried to write down a rigorous proof,
since i didn't know the small oh notation is available here and i was worried about the
fact that log x does not exist for x->infty, so...

Your answer helped me a lot. tks again

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