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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!

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4 Answers 4

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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 $x=a+\frac{2\log a}{\varepsilon}$.

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  • $\begingroup$ Thanks, very clear and easy to understand, and the specification of when exactly $\epsilon x$ overtakes $\ln(x)$ was particularly useful & insightful. $\endgroup$ Dec 10, 2012 at 1:28
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    $\begingroup$ @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. $\endgroup$ Dec 10, 2012 at 13:13
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For all $\varepsilon>0$ using L'Hospital's rule $$\lim\limits_{x \to +\infty} {\dfrac{\varepsilon x}{\ln{x}}}=\varepsilon \lim\limits_{x \to +\infty} {\dfrac{x}{\ln{x}}}=\varepsilon \lim\limits_{x \to +\infty} {\dfrac{1}{\frac{1}{x}}}=+\infty.$$

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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.

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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 $(e,\infty)$.

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