does the logic here seem correct Prove the following statement:
$$\frac{1}{x}<\ln(x)-\ln(x-1)<\frac{1}{x-1}$$
Proof:
$$\frac{-1}{x^2}<\frac{1}{x(x-1)}<\frac{-1}{(x-1)^2}$$
$$e^{(\frac{-1}{x^2})}<e^{(\frac{-1}{x(x-1)})}<e^{(\frac{-1}{(x-1)^2})}$$
$$\lim_{x\to\infty}e^{(\frac{-1}{x^2})}<\lim_{x\to\infty}e^{(\frac{-1}{x(x-1)})}<\lim_{x\to\infty}e^{(\frac{-1}{(x-1)^2})}$$
$$e^{0}<e^{0}<e^{0}$$
$$1<1<1$$
therefore MVT and we get the statement to be proven.
does anyone agree with me in the way i choose to prove the above statement? any feedback would be good thank you in advance!
 A: There's a mistake in your calculus: if we have a strictly inequality
$$f(x)<g(x)$$
then by passing to the limit we have 
$$\lim_{x\to a}f(x)\leq \lim_{x\to a}g(x),\qquad a\in \mathbb{R}\cup\{\infty\}$$
A: You've made this much too difficult, and your proof is a bit shady.
Try a different tact:
Note that for any $x>1$,
$$
\int_{x-1}^{x}\frac{1}{t}\,dt=\ln(x)-\ln(x-1).
$$
However, $\frac{1}{t}$ is monotone decreasing on $[x-1,x]$; can you come up with a way to use this to bound the above integral?
A: By the nature of question we must have $x > 1$. Now use mean value theorem to get $$\ln(x) - \ln(x - 1) = \{x - (x - 1)\}\cdot\dfrac{1}{y} = \dfrac{1}{y}$$ for some $x - 1 < y < x$. Thus we get $$\frac{1}{x} < \frac{1}{y} < \frac{1}{x - 1}$$ so that $$\frac{1}{x} < \ln x - \ln (x - 1) < \frac{1}{x - 1}$$
Regarding your approach I am not sure why you have differentiated the inequality in question. And the new identity obtained by differentiating is wrong because terms on left and right are negative and middle one is positive. Whenever you see expression like $f(a) - f(b)$ one possible technique is to apply the mean value theorem directly : $f(a) - f(b) = (a - b)f'(c)$ for some $a < c < b$.
