<s>[INST] 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.
 [/INST] Assume $n>0$.
By mean value theorem, there exists $c\in(n,n+1)$ such that
$$\frac{1}{n}-\frac{1}{x}=\frac{1}{c^2}(x-n)$$
Since $c>n$, we get $c^2>n^2$. So 
$$\left|\frac{1}{n}-\frac{1}{x}\right|<\frac{x-n}{n^2}\le\frac{1}{n^2}.$$
</s>