Evaluate L = lim as x to 0 (3/(3x^4-x^6)- 1/(x³arctan x))

Evaluate $\displaystyle L = \lim_{x\to0}\left(\frac{3}{3x^{4}-x^{6}}-\frac{1}{x^{3}\arctan x}\right)$.

Step 1: Identify the orders

\[ \frac{3}{3x^{4}-x^{6}} = \frac{3}{x^{4}\bigl(3-x^{2}\bigr)} = \frac{1}{x^{4}}\cdot\frac{3}{3-x^{2}}, \tag{1} \]

and since $\arctan x\sim x$, the second term behaves like $\dfrac{1}{x^{4}}$ as well. Both are of order $x^{-4}$, so the difference is $\infty-\infty$.

Step 2: Combine over a common denominator

\[ \frac{3}{3x^{4}-x^{6}}-\frac{1}{x^{3}\arctan x} = \frac{3\arctan x-x\bigl(3-x^{2}\bigr)}{x^{4}\bigl(3-x^{2}\bigr)\arctan x} = \frac{3\arctan x-3x+x^{3}}{x^{4}\bigl(3-x^{2}\bigr)\arctan x}. \tag{2} \]

Step 3: Expand the numerator

From $\arctan x = x-\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}-\cdots$, $3\arctan x = 3x-x^{3}+\frac{3x^{5}}{5}+O\!\left(x^{7}\right)$.

\[ 3\arctan x-3x+x^{3} = \left(3x-x^{3}+\frac{3x^{5}}{5}\right)-3x+x^{3}+O\!\left(x^{7}\right) = \frac{3x^{5}}{5}+O\!\left(x^{7}\right). \tag{3} \]

Both the linear and the cubic terms cancel exactly — that double cancellation is the point of the problem — leaving a numerator of order $x^{5}$.

Step 4: Expand the denominator

\[ x^{4}\bigl(3-x^{2}\bigr)\arctan x = x^{4}\cdot3\left(1-\frac{x^{2}}{3}\right)\cdot x\left(1-\frac{x^{2}}{3}+O\!\left(x^{4}\right)\right) = 3x^{5}\left(1-\frac{2x^{2}}{3}+O\!\left(x^{4}\right)\right). \tag{4} \]

Step 5: Form the ratio and take the limit

\[ \frac{\dfrac{3x^{5}}{5}+O\!\left(x^{7}\right)}{3x^{5}\left(1-\dfrac{2x^{2}}{3}+\cdots\right)} = \frac{\dfrac35+O\!\left(x^{2}\right)}{3\left(1-\dfrac{2x^{2}}{3}+\cdots\right)} \longrightarrow \frac{3/5}{3} = \frac15. \]

Step 6: Verification

\[x = 0.1:\ 0.1999090; \qquad x = 0.05:\ 0.1999765; \qquad x = 0.01:\ 0.1999990 \;\;\checkmark\]

\[\text{The full expansion is } \frac15-\frac{x^{2}}{105}+\cdots:\ \text{at } x = 0.1,\ 0.2-0.0000952 = 0.1999048 \;\;\checkmark\]

\[\text{Even function: } x = -0.1 \text{ gives the same value} \;\;\checkmark\]