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!

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