Prove $f$ is not of bounded variation function. Let $f(x)=\dfrac{1}{x}$, for $0<x\leq 1$ and $f(x)=0$, for $x=0$. Prove $f$ is not of bounded variation function.
This is my attemp.
\begin{align}
V(f,[0,1])&=\sum\limits_{k=1}^{n} \left\vert f{(x_k)}-f{(x_{k-1})}\right\vert\\
&=\left\vert f{(x_1)}-f{(x_{0})}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert
+\ldots +\left\vert f{(x_n)}-f{(x_{n-1})}\right\vert
\\
&=\left\vert f{(x_1)}-f{(0)}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert
+\ldots +\left\vert f{(1)}-f{(x_{n-1})}\right\vert
\\
&= \left\vert\dfrac{1}{x_1}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\\
&= 1-0\\
&=1
\end{align}
Now I confused to prove $f$ is not of bounded variation function, because $V(f,[0,1])=1<M$, for some $M>0$.
So, anyone can correct my answer if my answer is incorrect?
 A: Bounded variation requires the bound to hold for every finite set $x_k$, not just for one set.
To find a set where the total variation is large, you need some $x_i$ to be positive but small.  In fact, you can do it with just one intermediate point. Let  $M$ be given (with $M>1$) and let $x_1=1/M$
$$|f(0)-f(x_1)|+|f(x_1)-f(1))| = M + (M-1)=2M-1> M$$
A: $$\pi=\{a=0\leq x_1<x_2<...<x_n\leq b=1\}$$take $x_1=\epsilon$ now
$$ \left\vert\dfrac{1}{x_1}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert=\\
 \left\vert\dfrac{1}{\epsilon}-0\right\vert
+\left\vert\dfrac{1}{x_2}-\dfrac{1}{\epsilon}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\geq \\
\left\vert\dfrac{1}{\epsilon}-0\right\vert
+\left\vert\dfrac{1}{1}-\dfrac{1}{\epsilon}\right\vert
+\ldots
+\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\geq \frac2{\epsilon}-1+|\frac1{x_2}-\frac1{x_3}|+...\\
\geq \frac2{\epsilon}-1\\
\to \infty$$
A: $|\frac{1}{x_1}-0|+|\frac{1}{x_2}-\frac{1}{x_1}|+... +|\frac{1}{x_n}-\frac{1}{x_{n-1}}|$ is not equal to $1$.  It is $[\frac{1}{x_1}-0]+[\frac{1}{x_1}-\frac{1}{x_2}]+... +[\frac{1}{x_{n-1}}-\frac{1}{x_n}]$. This works out to $\frac 2 {x_1} -1$ which is not bounded.
