Let $f(x)=e^{-x} \cos x$. Show that $f$ is of bounded variation on any compact interval. For each $k$ in $\mathbb{N}$, compute $V(f ; 0,2 k \pi)$ and show that $\{ V(f; 0, 2 k \pi): k$ in $\mathbb{N} \}$ is an increasing sequence that is bounded above. Find its limit.
Since the exponential function is an increasing function, and $cos x$ has a bounded derivative, both of them will be bounded variations on any compact interval, along with their additions and multiplications. However, computing precisely the supremum of $\Delta f(x_j)$ summations for any partition of $[0; 2 \pi k]$ seems a little bit vague. Are there any hidden tricks to evaluate the total variation on $[0; 2 \pi k]$? Proving monotonicity and boundedness would be simple afterward
Note that total variation is additive in the sense that $V(f;a,c)+V(f;c,b) = V(f;a,b) $ and it follows that $$V(f; 0, 2k\pi) = \sum_{j=0}^{k-1} V(f; 2j\pi, 2j\pi + 2\pi) $$
For $x \in [2j\pi, 2j\pi + 2\pi]$ we have $f(x) = f(2j\pi +u) = e^{-2j\pi}e^{-u} \cos (2j\pi +u) =e^{-2j\pi}g(u)$ where $g(u) = e^{-u} \cos u$ for $u \in [0,2\pi]$.
Hence, $ V(f; 2j\pi, 2j\pi + 2\pi)= e^{-2j\pi}V(g,0,2\pi)$ and
$$V(f; 0, 2k\pi) = \sum_{j=0}^{k-1}e^{-2j\pi} V(g; 0, 2\pi) = V(g; 0, 2\pi)\sum_{j=0}^{k-1}e^{-2j\pi} = V(g; 0, 2\pi)\frac{1- e^{-2k\pi}}{1- e^{-2\pi}}$$
Thus, $\{V(f;0,2k\pi)\}_k$ is an increasing sequence with $V(f; 0, 2k\pi) \nearrow V(g;0,2\pi)/(1 - e^{-2\pi})$ as $k \to \infty$.
It is easy to compute $V(g;0,2\pi)$ since $g$ is monotonic on the intervals $\left[0,\frac{3\pi}{4}\right] $,$\left[\frac{3\pi}{4},\frac{7\pi}{4}\right] $ and $\left[\frac{7\pi}{4},2\pi\right] $.