Asymptotic expansion of roots of function Find expansions for all roots of the equations below as
$\epsilon \to 0$ with two nonzero terms in each expansion

(b) $\tan(x) = \frac{\epsilon}{x}$
Hint: Draw a graph to see where the roots are. You may need to consider the first root separately

I don't see how drawing the graph will help. Also how do I go about balancing the sizes of the roots?
Any help would be greatly appreciated.
Thank you!
 A: Yes, drawing a graph will help you see the following.
For the first zero, note that the min of $y=\epsilon/x$ is at the point $\left ( \sqrt{\epsilon},\sqrt{\epsilon} \right )$.  Note also that, for small $x$, $\tan{x} \approx x$.  Thus, for the first zero, when all quantities are small, the tangent curve will intersect the min of the hyperbola.  Thus, the first root is at approximately $x_1 = \arctan{\sqrt{\epsilon}} = \epsilon^{1/2} - \delta$, where $\delta \gt 0$ is a correction.  We find the correction by expanding the tangent function in its Taylor series as follows:
$$\tan{\sqrt{\epsilon}} \approx \sqrt{\epsilon} - \delta + \frac13 \left (\sqrt{\epsilon} - \delta \right )^3 \approx \sqrt{\epsilon} + \frac13 \epsilon^{3/2} - \delta $$
We set this equal to $\epsilon/(\sqrt{\epsilon}-\delta)$, and expanding for small $\delta$, we have
$$\sqrt{\epsilon} + \frac13 \epsilon^{3/2} - \delta \approx \sqrt{\epsilon} + \delta$$
or
$$x_1 = \sqrt{\epsilon}-\delta \approx \sqrt{\epsilon} - \frac16 \epsilon^{3/2}$$
For subsequent roots, the hyperbola intersects the tangent curves near the zero of the tangent at $x_n = n \pi$ for $n \ge 2$.  The correction $\delta$ satisfies
$$\delta (\delta + n \pi) = \epsilon$$
The reason for this is, again, the tangent curve is approximately the line $y=x-n \pi$ near the root $x_n$, so when $x_n = n \pi + \delta$, we have $\delta = \epsilon/(n \pi+\delta)$.  Thus, for small $\epsilon$, we have
$$x_n = n \pi + \frac{\epsilon}{n \pi}$$
