Check existence of limit $$
\lim_{k\rightarrow\infty} \int_o^1 |\cos{(kx)}|\,dx
$$
$$
\int_o^1 |cos(kx)|\,dx = \frac{1}{k}\int_o^k |\cos{y}|\,dy =\frac{f(k)}{k} 
$$
where I used $y=kx$ and $f(x)=\int_0^x |\cos{y}|\,dy$.
How can I proceed or what would be the easier approach?

Your answer is not true. Since 
$$\int_0^1 |\cos{(kx)}|dx = \frac{1}{k} \int_0^k |\cos{y}|dy.$$
Note that $$\dfrac2k\lfloor \frac{k}{\pi} \rfloor=\frac{1}{k} \int_0^{\lfloor k/\pi\rfloor\pi} |\cos{y}|dy\le\frac{1}{k} \int_0^k |\cos{y}|dy\le \frac{1}{k} \int_0^{(\lfloor k/\pi\rfloor+1)\pi} |\cos{y}|dy=\frac{2}{k}
\left(\lfloor \frac{k}{\pi} \rfloor+1\right)$$ 
This inequality follows from that we integrate via the first $(\lfloor k/\pi\rfloor+1)$ intervals ( i.e. $[0,(\lfloor k/\pi\rfloor+1)\pi]$ to get the right inequality since $k\le (\lfloor k/\pi\rfloor+1)\pi$. The left inequality follows the same way).Then by inequality we know that
$$\lim_{k\to\infty} \int_0^1 |\cos{(kx)}|dx=\frac{2}{\pi}$$
