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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.

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

1 Answer 1

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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.)

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

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  • $\begingroup$ Does anyone know how to use spoilers (the ">!") with multiine $\LaTeX$ equations? $\endgroup$
    – Clement C.
    Commented Oct 10, 2015 at 23:27
  • $\begingroup$ how do you get the series represented e^(-2kx) in to a function? $\endgroup$
    – spatel
    Commented Oct 10, 2015 at 23:42
  • $\begingroup$ @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 $e^{-kx} = (e^{-x})^k$. $\endgroup$
    – Clement C.
    Commented Oct 11, 2015 at 0:07

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