# Evaluate the following limit using Taylor series.

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.

• Can you please use LaTeX to type your limit expression? It is unclear what exactly is supposed to be 'over x^5.' Commented Oct 10, 2015 at 23:18
• Thank you Clement C. for formatting this correctly. Commented Oct 10, 2015 at 23:39

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}
• Does anyone know how to use spoilers (the ">!") with multiine $\LaTeX$ equations? Commented Oct 10, 2015 at 23:27
• @spatel This is an entirely different question, but if you know how to compute $\sum_{k=0}^\infty a^k$ when $a > 0$ (which is only guaranteed to exist for $0 < a < 1$), then you can use the fact that $e^{-kx} = (e^{-x})^k$. Commented Oct 11, 2015 at 0:07