TITLE: Why does $\ln(x) = \epsilon x$ have 2 solutions?
QUESTION [7 upvotes]: 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!

REPLY [5 votes]: 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 $x=a+\frac{2\log a}{\varepsilon}$.

REPLY [2 votes]: For all $\varepsilon>0$ using L'Hospital's rule 
$$\lim\limits_{x \to +\infty} {\dfrac{\varepsilon x}{\ln{x}}}=\varepsilon \lim\limits_{x \to +\infty} {\dfrac{x}{\ln{x}}}=\varepsilon \lim\limits_{x \to +\infty} {\dfrac{1}{\frac{1}{x}}}=+\infty.$$

REPLY [2 votes]: 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.