Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ for all $n \le x \le n+1$, using the mean value theorem applied to $f(x) = \frac{1}{x}$ 
Immediately, I can recognize some components of the mean value theorem. $\frac{1}{n^2}$ likely comes from the slope of $f$ at $n$, and the  $|\frac{1}{n} - \frac{1}{x}|$ expression likely comes from the secant expression $\frac{\frac{1}{n} - \frac{1}{x}}{x}$. But I cannot figure out how they fit together in the end.

If $x = n$, then the left hand side is $0$ and the result holds. So let's assume that $n < x \leq n+1$. Then applying the mean-value theorem to the function $f(t) \colon= 1/t$ for $t \in [n, x]$, we find that there is a real number $c \in (n,x)$ such that 
$$ f(x)-f(n) = f\prime(c) \cdot (x-n) $$
or 
$$\frac{1}{x}-\frac{1}{n} = - \frac{1}{c^2} \cdot (x-n). $$
So 
$$ |\frac{1}{n}-\frac{1}{x}| =  \frac{1}{n}-\frac{1}{x} = \frac{1}{c^2} \cdot (x-n) \leq \frac{1}{c^2} < \frac{1}{n^2}. $$
We of course assume that $n$ is positive. 
