Rasine:
I'll do this one for you, so that you can see how we do these problems.
Now, one of the reasons why I substituted 1+\delta at the variable's place, is that we have the trivial relations:
Take any x so that 1\leq{x}<{1}+\delta, and define \delta_{x}=x-1:
\delta_{x}<\delta, \delta_{x}^{2}<\delta^{2},\sqrt{\delta_{x}}<\sqrt{\delta}
Therefore, 0\leq{f(x)}<f(1+\delta)
Furthermore, with 0 being the limit of f at x=1, we have that:
|f(x)-0|<|f(1+\delta)-0|=(\delta^{2}+3\delta+3)\sqrt{\delta}
Thus, if we can assign a value of delta so that (\delta^{2}+3\delta+3)\sqrt{\delta}<0.7, then that inequality holds for any choice of x lying between 1 and 1+\delta, and our proof is finished.
Now, how do we find such a workable delta value.
There many ways of doing this, here's perhaps the simplest one:
If we ASSUME that \delta\leq{1}, then we have:
\delta^{2}+3\delta+3<1^{2}+3*1+3=7
Hence, we have:
(\delta^{2}+3\delta+3)\sqrt{\delta}<7\sqrt{\delta}, \delta<1
Now, can we make 7\sqrt{\delta}\leq{0.7}?
Indeed we can, if we set \delta\leq{0.01}
But, therefore, since 0.01<1, it follows that by choosing \delta=0.01, we have the inequality sequence, for every x 1<{x}<{1+\delta},\delta=0.01:
|f(x)-0|<(\delta^{2}+3\delta+3)\sqrt{\delta}<7\sqrt{\delta}<7*0.1=0.7
which is our desired result.
