How to use mean vaule theorem here

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SUMMARY

The discussion centers on the application of the Mean Value Theorem (MVT) to prove that if a function f(x) is differentiable on (a, +∞) and lim [f(x)]/x = 0 as x approaches +∞, then lim inf |f'(x)| = 0 as x approaches +∞. The solution provided involves analyzing the limit of f(2m)/m and applying the MVT to establish that f'(c) = 0 for some c in the interval [x, 2x]. However, the original poster expresses confusion regarding the proof and the specific value of c.

PREREQUISITES
  • Understanding of the Mean Value Theorem (MVT)
  • Knowledge of limits and their properties
  • Familiarity with differentiable functions
  • Basic calculus concepts, particularly limits at infinity
NEXT STEPS
  • Study the Mean Value Theorem in detail, focusing on its implications for differentiable functions.
  • Learn about the concept of lim inf and how it applies to sequences and functions.
  • Explore examples of functions that satisfy the conditions of the problem to understand the behavior of their derivatives.
  • Investigate counterexamples where lim inf |f'(x)| does not equal 0 despite lim [f(x)]/x = 0.
USEFUL FOR

Mathematics students, calculus instructors, and anyone studying real analysis or the properties of differentiable functions will benefit from this discussion.

transgalactic
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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

??
 
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transgalactic said:
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

??
Thanks for including your work in your post!

Regarding your question about the Mean Value Theorem:
The theorem says that if f is continuous on [a, b], and differentiable on (a, b), then there exists some number c in (a, b) such that
[tex]f'(c) = \frac{f(b) - f(a)}{b - a}[/tex]

The MVT doesn't tell you the value of c or how to find it; it just says that such a number exists.

I believe that what you're reading is saying that since f'(c) = 0 for some number c in [x, 2x], then this is also true in the limit as x approaches infinity.
 
so how to prove that
lim inf |f'(x)|=0 as x->+infinity
 

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