What is the correct method of finding the leading order behavior of a function in a given limit?

Question

I am kind of confused about finding the correct leading order behavior of a function. Example: If I want to understand the behavior of the following function

$$f(x)=\coth(x)-\frac{1}{x}$$

I can write it as (after some algebraic manipulations)

$$f(x)= 1+ \frac{2}{e^{2x}-1} - \frac{1}{x}$$

I can series expand some of the terms to have

$$f(x)= 1+ \frac{2}{2x+2x^2+\frac{4}{3}x^3+...} -\frac{1}{x}$$

Now, I want the leading order behavior of this function in the limit $x \rightarrow 0$. The correct answer is that it goes as $\frac{x}{3}$ in this limit. And, $f(0) = 0$. But I do not know how ?.

My confusion is general and I want to know, in general, how should I approach a problem like this. What is the best way to expand a function like this. I am a physics student so please excuse my ignorance.

Thanks

Answer 1

HINT:

Write the hyperbolic cotangent as

$$\coth(x)=\frac{1+\frac12x^2+O(x^4)}{x(1+\frac16x^2+O(x^4))}=\frac1x\left(\frac{1+\frac12x^2+O(x^4)}{1+\frac16x^2+O(x^4)}\right)\tag 1$$

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SPOILER ALERT Scroll over the highlighted area to reveal solution

Starting with the right-hand side of $(1)$, we have $$\frac1x\left(\frac{1+\frac12x^2+O(x^4)}{1+\frac16x^2+O(x^4)}\right)=\frac1x\left(1+\frac13x^2+O(x^4)\right)$$Finally, we have $$\coth(x)-\frac1x=\frac13x+O(x^3)$$And we are done!

Answer 2

Differentiating (and, optionally, taking advantage of the fact that $\tanh x$ is odd) gives that $$\tanh x = x - \frac{x^3}{3} + O(x^5) = x\left(1 - \frac{x^2}{3} + O(x^4)\right),$$ and so $$\coth x = \frac{1}{\tanh x} = \frac{1}{x} \left(1 + \frac{x^2}{3} + O(x^4)\right) = \frac{1}{x} + \frac{x}{3} + O(x^3) .$$ Rearranging gives $$\color{#bf0000}{\boxed{\coth x - \frac{1}{x} = \frac{x}{3} + O(x^3)}}.$$