- #1
transgalactic
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there is a function f(x) which is differentiable on (a,+infinity)
suppose lim [f(x)]/x =0 as x->+infinity
prove that lim inf |f'(x)|=0 as x->+infinity ?
does this expression lim inf f'(x)=0 has to be true
if not
present a disproving example
?
i got this solution :
First consider [itex]\lim_{m\to\infty}\frac{f(2m)}{m}[/itex], let [itex]2m=x[/itex] and this limit becomes [itex]2\lim_{x\to\infty}\frac{f(x)}{x}=0[/itex]. So [itex]\lim_{x\to\infty}\frac{f(2x)}{x}[/itex] exists and equals [itex]0[/itex]. So [itex]\lim_{x\to\infty}\frac{f(x)}{x}=0\implies \lim_{x\to\infty}\frac{f(2x)-f(x)}{x}=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0[/itex]. So now consider the interval [itex][x,2x][/itex] and apply the MVT letting x approach infinity.
I tried to follow your logic.
and i got a few question regarding it??
http://img82.imageshack.us/my.php?image=14440292jz6.gif
suppose lim [f(x)]/x =0 as x->+infinity
prove that lim inf |f'(x)|=0 as x->+infinity ?
does this expression lim inf f'(x)=0 has to be true
if not
present a disproving example
?
i got this solution :
First consider [itex]\lim_{m\to\infty}\frac{f(2m)}{m}[/itex], let [itex]2m=x[/itex] and this limit becomes [itex]2\lim_{x\to\infty}\frac{f(x)}{x}=0[/itex]. So [itex]\lim_{x\to\infty}\frac{f(2x)}{x}[/itex] exists and equals [itex]0[/itex]. So [itex]\lim_{x\to\infty}\frac{f(x)}{x}=0\implies \lim_{x\to\infty}\frac{f(2x)-f(x)}{x}=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0[/itex]. So now consider the interval [itex][x,2x][/itex] and apply the MVT letting x approach infinity.
I tried to follow your logic.
and i got a few question regarding it??
http://img82.imageshack.us/my.php?image=14440292jz6.gif