prove or desprove by example

1.prove or desprove by example:
a)
$\displaystyle sup\{\frac{[x]}{[x]+1}|x\geq0\}=1$
b)prove or desprove that $\displaystyle x^{3}$ 'uniformly' continuous in R
c)if f(x) monotonicly decreasing in $\displaystyle [0,\infty)$ and $\displaystyle lim_{x->\infty}=0$then f(x)>0 for $\displaystyle x\in[0,\infty)$
d)f,g are continues in [a,b] if $\displaystyle x\in[a,b] \int_{a}^{x}f(t)dt=\int_{a}^{x}g(t)dt$ then f=g in [a,b]


regarding the first one i couldnt find a value for which the supremum is 1
for part a) i cant say if its correct or not

?

Quote:

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Originally Posted by transgalactic

1.prove or desprove by example:
a)
$\displaystyle sup\{\frac{[x]}{[x]+1}|x\geq0\}=1$
b)prove that $\displaystyle x^{3}$ 'uniformly' continuous in R
c)if f(x) monotonicly decreasing in $\displaystyle [0,\infty)$ and $\displaystyle lim_{x->\infty}=0$then f(x)>0 for $\displaystyle x\in[0,\infty)$
d)f,g are continues in [a,b] if $\displaystyle x\in[a,b] \int_{a}^{x}f(t)dt=\int_{a}^{x}g(t)dt$ then f=g in [a,b]


regarding the first one i couldnt find a value for which the supremum is 1

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That makes no sense. The supremum does not depend upon a "value". It should be obvious that 1 is an upper bound for this. Is there a smaller upper bound? That is, is there a number $\displaystyle \alpha< 1$ such that $\displaystyle \frac{x}{x+1}< \alpha$ for all x> 1? What do you get if you solve that inequality for x?

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for part a) i cant say if its correct or not

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ok so 1 is the upper bound and in order to prove that its supremum ,we need to show
that its the smallest upper bound.

suupose 1 is not the smallest and $\displaystyle sup\{\frac{[x]}{[x]+1}|x\geq0\}=t<1$.
now i think we should use $\displaystyle lim_{x->\infty}\frac{[x]}{[x]+1}$
but i cant use lhopital law here like normal variable because we have to get rid of [x].
fo x>0
x-1<[x]<x+1
so $\displaystyle lim_{x->\infty}\frac{[x]}{[x]+1} $>$\displaystyle lim_{x->\infty}\frac{x-1}{x+1+1}$
so i have the result of a smaller limit.
how i know the result of the bigger one
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Instead of using the L'Hopital's rule, I would follow the suggestion above and solve [x]/([x] + 1) < t for x. If [x] means the integer part of x, I would solve it when x is an integer.

so you want me to replace [x] with x,and prove that t is its supremum
?

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Originally Posted by transgalactic

so you want me to replace [x] with x,and prove that t is its supremum
?

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No,

Quote:

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Originally Posted by emakarov

I would follow the suggestion above and solve [x]/([x] + 1) < t for x

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I have not said anything about proving that t is a supremum. The suggestion is to solve the inequality on integers, so you can remove the square brackets.

ahhhh instead of R i solve it on N
but how to make the transition back to R
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but how to make the transition back to R
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Each integer is a real number.

We know that 1 is an upper bound of $\displaystyle \{[x]/([x]+1)\mid x\ge0\}$. The idea is to prove that anything less than 1 is not an upper bound. Suppose t < 1. Then t is not an upper bound iff the inequality $\displaystyle [x]/([x]+1) > t$ is true for some $\displaystyle x\ge0$. Well, the suspicion is that it is true for infinitely many $\displaystyle x\ge 0$, and even true for infinitely many positive integers. So, try to solve this inequality when x is an integer and see if this is true.

ahhh if i will prove for integer that $\displaystyle \frac{x}{x+1}>t$ then t cannot be an upper bound and specially supremum.
ok the integer group is not bounded from top so $\displaystyle N> \frac{x}{x+1}$
but i dont know how to continue and i dont know where to put "t"
i only remmember the part when N is not bounded and we put it to be larger then the fracture.

another way i remmber is
$\displaystyle lim_{x->\infty}\frac{x}{x+1}=1$
so for every $\displaystyle \epsilon >0$ there is $\displaystyle M >0$ for which every $\displaystyle x>M$
$\displaystyle |\frac{x}{x+1}-1|<\epsilon$
i can open this
to
$\displaystyle 1-\epsilon<\frac{x}{x+1}<\epsilon+1$
so i can choose $\displaystyle 1-\epsilon=t$
and
$\displaystyle t<\frac{x}{x+1}$
correct?

Quote:

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Originally Posted by transgalactic

ahhh if i will prove for integer that $\displaystyle \frac{x}{x+1}>t$ then t cannot be an upper bound and specially supremum.
ok the integer group is not bounded from top so $\displaystyle N> \frac{x}{x+1}$
but i dont know how to continue and i dont know where to put "t"

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You should review how to solve linear inequalities such as x / (x + 1) > t. Without it, frankly, talking about limits, continuity, or any other calculus topic looses meaning. (I am assuming this is the issue because this thread has four suggestions to solve this inequality, and my last post explains how this would help solve the problem.)

Quote:

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Originally Posted by transgalactic

another way i remmber is
$\displaystyle lim_{x->\infty}\frac{x}{x+1}=1$
so for every $\displaystyle \epsilon >0$ there is $\displaystyle M >0$ for which every $\displaystyle x>M$
$\displaystyle |\frac{x}{x+1}-1|<\epsilon$
i can open this
to
$\displaystyle 1-\epsilon<\frac{x}{x+1}<\epsilon+1$
so i can choose $\displaystyle 1-\epsilon=t$
and
$\displaystyle t<\frac{x}{x+1}$
correct?

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There is sense in this, but, again, I think it's meaningless doing limits without a solid grasp of algebra.

i know how to solve innequalitties
you didnt say which one.i have presented two ways of continuing this solution
could you say which one to follow
?
be more spesific

Quote:

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Originally Posted by emakarov

We know that 1 is an upper bound of $\displaystyle \{[x]/([x]+1)\mid x\ge0\}$. The idea is to prove that anything less than 1 is not an upper bound. Suppose t < 1. Then t is not an upper bound iff the inequality $\displaystyle [x]/([x]+1) > t$ is true for some $\displaystyle x\ge0$. Well, the suspicion is that it is true for infinitely many $\displaystyle x\ge 0$, and even true for infinitely many positive integers. So, try to solve this inequality when x is an integer and see if this is true.

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ahhhhhhhhhhhhh ok
x/(x+1) >t
x>t(x+1)
x-tx>1
x>1/(1-t)
0<t<1 from the data in the question
but how it proves that t is not upper bound
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Since 0<t<1, 1/(1-t) is well-defined positive number. You can take x to be any integer greater than 1/(1-t), and then [x] / ([x] + 1) > t, disproving that t is the upper bound of {[x] / ([x] + 1) | x >= 0}.

ahh i understand now part a

regarding part b:
i know a guide line that any curve above f(x)=x is not uniformly continues

so we must prove that there is $\displaystyle \epsilon>0$ so for all $\displaystyle \delta>0$
there is x,y in R which $\displaystyle |x-y|<\delta$ and$\displaystyle |f(x)-f(y)|=>\epsilon$

$\displaystyle |x^3-y^3|=|(x-y)(x^2+xy+y^2)|<|\delta (x^2+xy+y^2)|$
so i choose $\displaystyle \epsilon=2$
$\displaystyle x=\frac{2}{\sqrt{\delta}}$
$\displaystyle y=\frac{3}{\sqrt{\delta}}$
but i know only that $\displaystyle |x-y|<\delta$
i need to find some delta expression so$\displaystyle |x-y|$ will be greater that it.

so the result of the x,y value input will give use greater then 2.

but here i have smaller then delta only expression
what do?