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Find the limit as x goes to 0 of (arcsin x / x)1/x2

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}}. \]

First, we rewrite the expression using the definition of the exponential:

    \[ \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}} = e^{\frac{1}{x^2} \log \left(\frac{\arcsin x}{x} \right)}. \]

Next, we need to get a series expansion for \arcsin x as x \to 0. The most straightforward way is to take the first few derivatives (we’ll only need the x^3 term).

    \begin{align*} f(x) &= \arcsin x & \implies && f(0) &= 0 \\[9pt] f'(x) &= \frac{1}{\sqrt{1-x^2}} & \implies && f'(0) &= 1 \\[9pt] f''(x) &= \frac{-x}{(1-x^2)^{\frac{3}{2}}} & \implies && f''(0) &= 0 \\[9pt] f'''(x) &= \frac{1+2x^2}{(1-x^2)^{\frac{5}{2}}} & \implies && f'''(0) &= 1. \end{align*}

Therefore, we have as x \to 0

    \[ \arcsin x = x + \frac{x^3}{6} + o(x^4). \]

Hence, as x \to 0

    \[ \frac{\arcsin x}{x} = 1 + \frac{x^2}{6} + o(x^3). \]

Since this is going to 1 as x \to 0 we may apply this exercise (Section 7.11, Exercise #4) to conclude

    \begin{align*} \log \left( \frac{\arcsin x}{x} \right) &= \left( \frac{\arcsin x}{x} - 1\right) - \frac{1}{2} \left( \frac{\arcsin x}{x} - 1 \right)^2 + o \left( \left( \frac{\arcsin x}{x} - 1 \right)^2 \right) \\[9pt] &= \left( \frac{x^2}{6} + o(x^3) \right) - \frac{1}{2} \left( \frac{x^2}{6} + o(x^3) \right)^2 + o \left( \left( \frac{x^2}{6} + o(x^3) \right)^2 \right) \\[9pt] &= \frac{x^2}{6} + o (x^3). \end{align*}

Now, we have the expansion as x \to 0

    \[ \frac{1}{x^2}\log \left( \frac{\arcsin x}{x} \right) = \frac{1}{6} + o (x). \]

Therefore,

    \begin{align*} \lim_{x \to 0} \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}} &= \lim_{x \to 0} e^{ \frac{1}{x^2} \log \left( \frac{\arcsin x}{x} \right)} \\[9pt] &= e^{\frac{1}{6}} \lim_{x \to 0} e^{o(x)} \\ &= e^{\frac{1}{6}}. \end{align*}

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