$a)\frac {1}{x^2+1} $
$b)\cos^3x$
$c)\frac {x^2}{x^2+2} $
$d)x\sin x$
$a)|\frac{1}{x^2+1}-\frac{1}{y^2+1}|\leq|\frac{|x-y|(|x|+|y|)}{(1+x^2)(1+y^2)}|\leq \frac{|x-y||x|}{(1+x^2)(1+y^2)} +\frac{|x-y||y|}{(1+x^2)(1+y^2)}<\frac{|x-y|}{1+y^2}+\frac{|x-y|}{1+x^2}$ ( since $\frac{|y|}{1+y^2}<1$ and $\frac{|x|}{1+x^2}<1)$
Can we take $L=\frac{2+x^2+y^2}{(1+x^2)(1+y^2)}$ for lipschitz?
b)from the definition of u.continuty $|\cos^3x-\cos^3y|=|\cos x-\cos y||1+\cos x\cos y|<|1+\cos x \cos y|$
$\leq |1+\cos(x+y)+\cos(x-y)||\leq|2+\cos(x-y)|<|2+\delta)| ($ since $\cos(x-y)<|x-y|<\delta )$ so if we choose $\delta=\epsilon-2$. f(x) is uniformly continous?
$c)|\frac{x^2}{x^2+2}-\frac{y^2}{y^2+2}|<|\frac{2|x-y||x+y|}{(x^2+2)(y^2+2)}$ how can I continue?
d)we must show there is a delta for all $\epsilon>0$ , for $|x-y|<\delta$, $|x\sin x-x\sin y| <\epsilon$
but when we take $x=m+2k\pi $ and $y=n+2k\pi$ $|(m+2k\pi)(\sin(m+2k\pi))-(n+2k\pi) \sin(m+2k\pi)|$ so $|m\sin m-n\sin n+2k\pi(\sin n-\sin m)|$
if we choose k big enough this inequality can be greater than epsilon. so is not uniformly continous?
Hint If a function $f$ is lipschitzian then it's uniformly continuous and notice that if $f'$ is bounded then $f$ is lipschitzian. We can apply the above remark to a,b,c).
Now for d) let $x_n=2n\pi$ and $y_n=x_n+\frac\pi2$ then we have $$y_n-x_n=\frac\pi2\xrightarrow{n\to\infty}\frac\pi2$$ and $$f(y_n)-f(x_n)\xrightarrow{n\to\infty}\infty$$ so what we can conclude?