Find lim as x to 0 arcsin x ·cot x

Find $\displaystyle\lim_{x\to0}\arcsin x\cdot\cot x$.

Step 1: Identify the indeterminate form

As $x\to0$ we have $\arcsin x\to0$, while $\cot x = \dfrac{\cos x}{\sin x}$ becomes infinite — $+\infty$ from the right, $-\infty$ from the left. Either way the product is of the form $0\cdot\infty$.

Step 2: Rewrite as a fraction

\[ \arcsin x\cdot\cot x = \frac{\arcsin x\cdot\cos x}{\sin x}. \tag{1} \]

Since $\cos x\to1$, the behaviour is governed by $\dfrac{\arcsin x}{\sin x}$, and the sign trouble disappears: numerator and denominator change sign together.

Step 3: Apply the standard equivalences

As $x\to0$ both $\arcsin x\sim x$ and $\sin x\sim x$, so

\[ \frac{\arcsin x}{\sin x} \sim \frac{x}{x} = 1. \]

Combining with $\cos x\to1$ in (1): \[\lim_{x\to0}\frac{\arcsin x\cdot\cos x}{\sin x} = \lim_{x\to0}\frac{\arcsin x}{\sin x}\cdot\cos x = 1\cdot1 = 1.\]

Step 4: Verification

\[x = 0.1:\ 0.9983331; \qquad x = -0.1:\ 0.9983331; \qquad x = 0.01:\ 0.9999833 \;\;\checkmark\]

\[\text{By l'H\^opital on } \frac{\arcsin x}{\tan x}:\ \lim_{x\to0}\frac{1/\sqrt{1-x^{2}}}{1/\cos^{2}x} = \frac{1}{1} = 1 \;\;\checkmark\]

\[\text{Series: } \frac{\arcsin x}{\tan x} = \frac{x+\tfrac{x^{3}}{6}+\cdots}{x+\tfrac{x^{3}}{3}+\cdots} = 1-\frac{x^{2}}{6}+\cdots \;\;\checkmark\]