# Thread: how to show if these are uniformly continuous

1. ## how to show if these are uniformly continuous

$f(x) = xsinx, \ \ \ \ g(x)=e^{-x^{4}}$

2. Originally Posted by silversand
$f(x) = xsinx, \ \ \ \ g(x)=e^{-x^{4}}$

The first one is not uniformly continous( I don't think)

Here is number two

Since g(x) is differentable $g'(x)=-4x^3e^{-x^4}$ and

the derivative is bounded on all of $\mathbb{R}$.

so $f'(x) < M$ for some $M \in \mathbb{R}$

let $\epsilon > 0$ set $\delta=\frac{\epsilon}{M}$

So now if $|x-y|< \delta$

We need to show that $|f(x)-f(y)|< \epsilon$

Now by the Mean Value theorem on $[x,y]$

$f(x)-f(y)=f'(c)(x-y)$

$|f(x)-f(y)|=|f'(c)(x-y)|=|f'(c)||x-y|=f'(c)\cdot \delta =f'(c) \frac{\epsilon}{M}< \epsilon$

3. Originally Posted by TheEmptySet
The first one is not uniformly continous( I don't think)

Here is number two

Since g(x) is differentable $g'(x)=-4x^3e^{-x^4}$ and

the derivative is bounded on all of $\mathbb{R}$.

so $f'(x) < M$ for some $M \in \mathbb{R}$

let $\epsilon > 0$ set $\delta=\frac{\epsilon}{M}$

So now if $|x-y|< \delta$

We need to show that $|f(x)-f(y)|< \epsilon$

Now by the Mean Value theorem on $[x,y]$

$f(x)-f(y)=f'(c)(x-y)$

$|f(x)-f(y)|=|f'(c)(x-y)|=|f'(c)||x-y|=f'(c)\cdot \delta =f'(c) \frac{\epsilon}{M}< \epsilon$
the first one is not uniformly cont on $\mathbb{R}$ b/c the derivative is $xcos(x)+sin(x)(1)$. so $f' \rightarrow \infty$ as $x \rightarrow \infty$ contradicting the fact that the derivative must be bounded.