Answer
$\dfrac{3}{5}$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{3 \tan^{-1} x-3x+x^3}{x^5}$
Recall the series for $\tan^{-1}(x)=\Sigma_{k =0}^{\infty} (-1)^{k} \dfrac{(x)^{2k+1}}{2k+1}$
Now, $L=\lim\limits_{x \to 0} [\dfrac{3 \Sigma_{k =0}^{\infty} (-1)^{k} \dfrac{(x)^{2k+1}}{2k+1}}{x^5}-\dfrac{2}{x^4}+\dfrac{1}{x^2}]\\=\lim\limits_{x \to 0} [3 (-1)^{0} \dfrac{(x)^{2(0)-4}}{2(0)+1}+3 (-1)^{1} \dfrac{(x)^{2(1)-4}}{2(1)+1}+......]-\dfrac{3}{x^4}+\dfrac{1}{x^2}]\\=\lim\limits_{x \to 0} [\dfrac{3}{5}-\dfrac{3x^2}{7}+\dfrac{3x^4}{9}-.......]\\=\dfrac{3}{5}-\dfrac{3(0)^2}{7}+\dfrac{3(0)^4}{9}-.......]\\=\dfrac{3}{5}$
