Why does $\ln(x) = \epsilon x$ have 2 solutions? 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!

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 $(e,\infty)$.  
