7
$\begingroup$

I was working on a problem involving perturbation methods and it asked me to sketch the graph of $\ln(x) = \epsilon x$ and explain why it must have 2 solutions. Clearly there is a solution near $x=1$ which depends on the value of $\epsilon$, but I fail to see why there must be a solution near $x \rightarrow \infty$. It was my understanding that $\ln x$ has no horizontal asymptote and continues to grow indefinitely, where for really small values of $\epsilon, \epsilon x$ should grow incredibly slowly. How can I 'see' that there are two solutions?

Thanks!

$\endgroup$
5
$\begingroup$

The slope of $\log x$ falls of as $1/x$. So for, say, $x>1/\varepsilon$ the logarithm always grows strictly slower than $\varepsilon x$, and the latter will eventually overtake it.

More precisely, if we put $a=2/\varepsilon$ then at any point to the right of $a$, the function $\varepsilon x$ gains at least $\varepsilon/2$ on the logarithm for each unit increase in $x$. Therefore $\varepsilon x$ will be larger than $\log x$ no later than at $x=a+\frac{2\log a}{\varepsilon}$.

$\endgroup$
  • $\begingroup$ Thanks, very clear and easy to understand, and the specification of when exactly $\epsilon x$ overtakes $\ln(x)$ was particularly useful & insightful. $\endgroup$ – still_learning Dec 10 '12 at 1:28
  • 1
    $\begingroup$ @MSEoris, beware: It's not a computation of "when exactly $\epsilon x$ overtakes $\ln(x)$" -- its a computation of an upper bound for the point of overtaking. $\endgroup$ – hmakholm left over Monica Dec 10 '12 at 13:13
2
$\begingroup$

For all $\varepsilon>0$ using L'Hospital's rule $$\lim\limits_{x \to +\infty} {\dfrac{\varepsilon x}{\ln{x}}}=\varepsilon \lim\limits_{x \to +\infty} {\dfrac{x}{\ln{x}}}=\varepsilon \lim\limits_{x \to +\infty} {\dfrac{1}{\frac{1}{x}}}=+\infty.$$

$\endgroup$
2
$\begingroup$

Right after the first solution, you have that $\ln x>\epsilon x \Rightarrow \ln x-\epsilon x>0$. However, $$\lim_{x\to\infty}(\ln x-\epsilon x)=\lim_{x\to\infty}\ln\frac{x}{e^{\epsilon x}}\to-\infty $$ since $\frac{x}{e^{\epsilon x}}\to 0$. (Assuming $\epsilon>0$)

Hence, $\ln x-\epsilon x$, must cross the $x$-axis a second time which gives your second solution.

$\endgroup$
1
$\begingroup$

Put $$f(x) = {\ln(x)\over x}.$$
Then you have $$f'(x) = {1 - \ln(x)\over x^2}.$$ When $x > e$, the function $f$ decreases; in fact is is easy to see it decreases to zero as $x\to\infty$. When $x < e$, $f$ increases. You can see that it has the $y$-axis as a vertical asymptote.

The graph has a global maximum at $x = e$; $f(e) = 1/e$. Note that this function is defined only when $x > 0$.

Now let $0 <\lambda < 1/e$. The graph will strike the horizontal line $y = \lambda$ once on $(0,e)$ and once on $(e,\infty)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.