Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$

Question

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.

Answer 1

$$f(x)=\frac{1}{x}\,\,\,\text{on the interval}\,\,\,[n\,,\,x]\,\,,\,n\leq x\leq n+1\Longrightarrow$$

$$\frac{f(x)-f(n)}{x-n}=f'(c)\,\,,\,\,\text{for some}\,\,\,c\in(n\,,\,x)\Longleftrightarrow$$

$$\frac{1}{x}-\frac{1}{n}=-\frac{1}{c^2}(x-n)$$

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

$$\left|\frac{1}{x}-\frac{1}{n}\right|\leq\frac{1}{c^2}<\frac{1}{n^2}\,\,,\,\,\text{since}\,\,\, c>n\,$$

Answer 2

$n-x\leq1 $ therefore $$|\frac{1}{n} - \frac{1}{x}| =|\frac{n-x}{nx}|\leq|\frac{1}{nx}|$$ i know $n \le x $ then $$\leq|\frac{1}{nx}|\leq\frac{1}{n^2}$$

Answer 3

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 $$\left|\frac{1}{n}-\frac{1}{x}\right|<\frac{x-n}{n^2}\le\frac{1}{n^2}.$$

Answer 4

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.