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Given that $\epsilon > 0$ prove that $\ln(x_2) - \ln(x_1) < \epsilon(x_2 - x_1)$ if $1 < x_1 < x_2$ and $x_1$ is sufficiently large. How large must $x_1$ be for the inequality to be guaranteed true?

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2 Answers 2

up vote 1 down vote accepted

It was a long time ago, but I couldn't restrain myself)

Solution 1 (without MVT):

Denote $f_1 (x) = \log x, \ f_2 (x) = \epsilon x$. At the lower bound $f_1(1)=0 < f_2(1)= \epsilon$. Hence, $f_2(1) >f_1(1)$. Taking the derivatives of both functions we see that the rate of growth of $f_{2}(x) >f_{1}(x)$ for $\delta > \frac{1}{\epsilon}$, which certainly includes $x_2$.

Solution 2 (with MVT):

Consider the function $h(x)= \epsilon x -\log x$. Then

$$ \frac{h(x_2)-h(x_1)}{x_2-x_1} = \frac{\epsilon(x_2 - x_1) - \log x_2 + \log x_1}{x_2 - x_1}= \epsilon -\frac{\log x_2 - \log x_1}{x_2 -x_1}=\epsilon - \frac{1}{\xi} $$ with $x_1 \leq \xi \leq x_2$. The last step is due to MVT and the derivative of the $\log$ function. Now we want this to be positive so that $h(x) >0$, so this holds for $\xi >\frac{1}{\epsilon}$, so the smallest possible value for $x_1$ is again $\frac{1}{\epsilon}$.

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Haha, thanks, Alex!!! –  Statz Feb 18 '13 at 21:56

Hint: What does the mean value theorem tell you? The derivative of $ln(x)$ is what? When is $1/y$ smaller than epsilon? Write everything out explicitly and I am sure you will get it.

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