# Thread: Uniform continuity help

1. ## Uniform continuity help

I have to detemine if these functions are uniformly continuous on the given intervals

1) f(x) = ln X on (0,1)
my solution :
Let ε >0 , we can let δ=ε
by the mean value theorm says there is c belongs to (0,1) such that
ln(X)-ln(Y)\ x-y =1\c <= 1
thus, whenever x-y < δ
So , we have ln(x) - ln(y)<= x-y< δ=ε
so the function is uniformly continuous on the interval .

Is my solution right ???

2)x\tan x on the interval (- infinity, infinity)
my solution :
tan-1(x) = cotan(x), it's not uniformly continuous and becomes infinitely large at the points pik, k = 0,1,2,3, x/tan(x) and not uniformly continuous at given interval.

I do not know how can i prove this by the definition . can anybody help me???

3)f(x)= cos(lnx) on the interval (0,1)
my solution:
f(x) is continuous on the interval (0,1)
for any two points from the interval (0,1) we can write the following expression:
|f(x1) - f(x2)| = |cos(ln(x1) - cos(ln(x2)| = |- 1/2 sin (lnx1 + lnx2)/2 sin (lnx1 - lnx2)/2 | = = 1/2 |sin(ln(x1x2)/2) sin (ln(x1/x2)/2)| <= 1/2, so the assumption of uniformly continuity is performed at this interval.

Is my solution right ?

4)f(x)=x2 tan-1x [0, infinity)
My solution :
I think it is the same on the first question
it's not continuous and becomes infinitely large at the points pik, k = 0,1,2,3, x/tan(x) and not continuous at given interval.

5)ex
on the interval (0,infinity)
my solution :
it is not uniformly continuous function , but i do not know how i can prove this by the definition

Please can anybody help me to solve these questions ???

2. Originally Posted by mariama
I have to detemine if these functions are uniformly continuous on the given intervals

1) f(x) = ln X on (0,1)
my solution :
Let ε >0 , we can let δ=ε
by the mean value theorm says there is c belongs to (0,1) such that
ln(X)-ln(Y)\ x-y =1\c <= 1
thus, whenever x-y < δ
So , we have ln(x) - ln(y)<= x-y< δ=ε
so the function is uniformly continuous on the interval .

Is my solution right ???
No because you can't say that $\displaystyle \frac 1c \leq 1$ whereas $\displaystyle 0<c\leq 1$.

3. Originally Posted by girdav
No because you can't say that $\displaystyle \frac 1c \leq 1$ whereas $\displaystyle 0<c\leq 1$.
So , how can i do it ?

4. You can't show the function is uniformly continuous because it is not the case. Let $\displaystyle x_n=e^{-n}$ and $\displaystyle y_n =2e^{-n}$. We have $\displaystyle |x_n-y_n| =e^{-n}$ and what about $\displaystyle |\ln x_n-\ln y_n|$ ?

5. Originally Posted by girdav
You can't show the function is uniformly continuous because it is not the case. Let $\displaystyle x_n=e^{-n}$ and $\displaystyle y_n =2e^{-n}$. We have $\displaystyle |x_n-y_n| =e^{-n}$ and what about $\displaystyle |\ln x_n-\ln y_n|$ ?
So, we have to approve that Lim | f(x+h)-f(x) |= infinty when x goes to a , and x goes to b

6. We don't now what a and b are.
Why don't you try what I proposed?

7. Originally Posted by girdav
We don't now what a and b are.
Why don't you try what I proposed?
I mean that , if this function is not uniformly continuous , so we have to prove that, not by defenition.
We have to approve that Lim when x goes to a and Lim when x goes to b for f(x+h)-f(x) = inifinty . where a, b the end points of the interval (a,b)

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