Evaluate the following limit using Taylor series.

Question

What is the limit, when $x\to0$, of $$\frac{4\tan x - 4x -\frac{4}{3}x^3}{x^5}?$$

I'm not sure how to expand this using the Taylor series.

Answer 1

Hint: the only thing to expand is the only thing that can be expanded using Taylor series/approximations. I.e., the $\tan$: everything else is already a polynomial.

Now, recall that $\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + o(x^5)$. Plugging it in the expression will give you the limit. (See below for more details.)

$$\begin{align} \frac{4\tan x - 4x -\frac{4}{3}x^3}{x^5} &= \frac{4(x + \frac{x^3}{3} + \frac{2x^5}{15}+o(x^5)) - 4x -\frac{4}{3}x^3}{x^5} \\ &= \frac{\frac{8x^5}{15}+o(x^5)}{x^5}\\ &= \frac{8}{15}+o(1). \end{align}$$