Darboux's sums of $f(x)=\frac{1}{x}$ Consider the function $f(x)=\frac{1}{x}$ on $[1,2]$. 

How to find $\epsilon&gt;0$, s.t for all partition $P$ of $[1,2]$ with $\lambda(P)&lt;\epsilon$, we will have that

$$|U_{f,P}-L_{f,P}|&lt;0.01$$

Where $U_{f,P}$ is upper Darboux's sum, and $L_{f,P}$ lower Darboux's sum.
You have that $$f(x)=\frac 1 x$$ is uniformly continuous over $[1,2]$, because it is continuous on $[1,2]$. Thus, given $\delta &gt;0$, pick $\rho&gt;0$ such that whenever $$|x-y|&lt;\rho$$ we have $$|f(x)-f(y)|&lt;\delta$$ for any choice of $x,y$. 

This means in particular that if $$\lambda(P)&lt;\rho$$ then $$\sup_{I_k} f-\inf_{I_k} f&lt;\delta$$ over each interval $I_k=[x_{k-1},x_k]$ of the partition. It follows that $$U(f,P)-L(f,P)&lt;\sum_{i=1}^n \delta (x_{i}-x_{i-1})=\delta (2-1)=\delta$$

So, it all reduces to finding this $\rho$. Your function is differentiable over $(1,2)$, and by the mean value theorem, we have that for each $x,y\in(1,2)$ there is a $c$ in between for which $$\left|\frac 1 x -\frac 1 y\right|=\frac{1}{c^2}|x-y|$$ But since $x,y&gt;1$, so is $c&gt;1$, so over $[1,2]$

$$\left|\frac 1 x -\frac 1 y\right|=\frac{1}{c^2}|x-y|&lt;|x-y|$$

Can you finish this off?
The function is decreasing, so the $sup$ and $inf$ over the $[p_i,p_{i+1}]$ are really easy to express.