Find lim as x to 0 (1/(x sin x)- 1/(x²))

Find $\displaystyle\lim_{x\to0}\left(\frac{1}{x\sin x}-\frac{1}{x^{2}}\right)$.

Step 1: Identify the indeterminate form

As $x\to0$, $x\sin x\sim x^{2}$, so both $\dfrac{1}{x\sin x}$ and $\dfrac{1}{x^{2}}$ tend to $+\infty$. The difference is of the form $\infty-\infty$.

Step 2: Combine into a single fraction

\[ \frac{1}{x\sin x}-\frac{1}{x^{2}} = \frac{x-\sin x}{x^{2}\sin x}. \tag{1} \]

At $x = 0$ numerator and denominator both vanish: the form is $\dfrac00$.

Step 3: Expand via the Maclaurin series

\[ \sin x = x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots \;\Longrightarrow\; x-\sin x = \frac{x^{3}}{6}-\frac{x^{5}}{120}+\cdots \tag{2} \]

\[ x^{2}\sin x = x^{2}\left(x-\frac{x^{3}}{6}+\cdots\right) = x^{3}-\frac{x^{5}}{6}+\cdots \tag{3} \]

Both are of order $x^{3}$ — the cancellation of the leading terms in (1) is exactly what makes the answer finite.

Step 4: Form the ratio

\[ \frac{x-\sin x}{x^{2}\sin x} = \frac{x^{3}\left(\dfrac16-\dfrac{x^{2}}{120}+\cdots\right)}{x^{3}\left(1-\dfrac{x^{2}}{6}+\cdots\right)} = \frac{\dfrac16-\dfrac{x^{2}}{120}+\cdots}{1-\dfrac{x^{2}}{6}+\cdots}. \]

Step 5: Evaluate the limit

\[ \lim_{x\to0}\frac{\dfrac16-\dfrac{x^{2}}{120}+\cdots}{1-\dfrac{x^{2}}{6}+\cdots} = \frac{1/6}{1} = \frac16. \]

Step 6: Verification

\[x = 0.1:\ 0.1668613; \qquad x = 0.01:\ 0.1666686; \qquad \tfrac16 = 0.1666667 \;\;\checkmark\]

\[\text{Dividing the series: } \frac16+\frac{7x^{2}}{360}+\cdots;\ \text{at } x = 0.1:\ 0.1666667+0.0001944 = 0.1668611 \;\;\checkmark\]

\[\text{Even function: } x = -0.1 \text{ gives the same } 0.1668613 \;\;\checkmark\]