transgalactic
- 1,386
- 0
i got this question:
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 was present whis this solution but i didnt quite understand it.
"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."
mean value theorem says [itex]f'(c)=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0[/itex]
i don't know what the value of c??
and it doesn't prove
lim inf |f'(x)|=0 as x->+infinity
??
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 was present whis this solution but i didnt quite understand it.
"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."
mean value theorem says [itex]f'(c)=\lim_{x\to\infty}\frac{f(2x)-f(x)}{2x-x}=0[/itex]
i don't know what the value of c??
and it doesn't prove
lim inf |f'(x)|=0 as x->+infinity
??