Limits Involving Trigonometric Functions I should prove using the limit definition that 
$$\lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) = 0.$$
I have a problem because the second function is  much too complex, so I think I need transformation.  And what form this function could have in case  I will transform it?

You can solve this problem with the Squeeze Theorem.
First, notice that $-1 \leq \cos(1/x) \leq 1$ (the cosine graph never goes beyond these bounds, no matter what you put inside as the argument).
Multiplying through by $x^{1/3}$, we get
$$
-x^{1/3} \leq x^{1/3}\cos(1/x) \leq x^{1/3}.
$$
Now, the Squeeze Theorem says
$$
\lim_{x \rightarrow 0} (-x^{1/3}) \leq \lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) \leq \lim_{x \rightarrow 0} \,  x^{1/3},
$$
so we investigate the left- and right-most limits.
Since, $x^{1/3}$ is continuous on $[0,\infty)$,
$$
\lim_{x \rightarrow 0} \,  x^{1/3} = \lim_{x \rightarrow 0} (-x^{1/3}) = 0.
$$
Finally, we have 
$$0 \leq \lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) \leq 0,$$
which forces us to conclude
$$
\lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) = 0.
$$
