<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] $$f(x)=\frac{1}{x}\,\,\,\text{on the interval}\,\,\,[n\,,\,x]\,\,,\,n\leq x\leq n+1\Longrightarrow$$
$$\frac{f(x)-f(n)}{x-n}=f'(c)\,\,,\,\,\text{for some}\,\,\,c\in(n\,,\,x)\Longleftrightarrow$$
$$\frac{1}{x}-\frac{1}{n}=-\frac{1}{c^2}(x-n)$$
But
$$\left|\frac{1}{c^2}(x-n)\right|\leq\frac{1}{c^2}\Longleftrightarrow |x-n|\leq 1$$
and the rightmost inequality is true by the given data (why?)  , thus we get
$$\left|\frac{1}{x}-\frac{1}{n}\right|\leq\frac{1}{c^2}<\frac{1}{n^2}\,\,,\,\,\text{since}\,\,\, c>n\,$$
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