# Uniformly Continuous.

• Mar 12th 2010, 07:41 AM
charikaar
Uniformly Continuous.
Prove or disapprove f(x) is uniformly continuous
f(x)=(sinx)/x for x not equal to zero.
and f(x)=1 for x=0

I know f(x) is uniformly continous but I can't prove it. how do i show the derivative $\displaystyle (xcosx-sinx)/x^{2}$ is bounded.

thanks
• Mar 12th 2010, 11:33 AM
1234567
Quote:

Originally Posted by charikaar
Prove or disapprove f(x) is uniformly continuous
f(x)=(sinx)/x for x not equal to zero.
and f(x)=1 for x=0

I know f(x) is uniformly continous but I can't prove it. how do i show the derivative $\displaystyle (xcosx-sinx)/x^{2}$ is bounded.

thanks

Show that the derivative is continous
• Mar 12th 2010, 11:42 AM
chisigma
Take into account that the coefficients of McLaurin expansion of $\displaystyle x\cdot \cos x - \sin x$ of degree < 3 are equal to zero, so that is $\displaystyle f^{'}(0)=0$ and that both the functions $\displaystyle \frac{\cos x}{x}$ and $\displaystyle \frac{\sin x}{x^{2}}$ tend to 0 if x tends to infinity...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$