Evaluating $\lim_{x\to {0}}\frac{1}{x\arcsin x} - \frac{1}{x^2}$ without L'Hôpital's rule

Question

So I have this limit ... $$\lim_{x\to {0}}\frac{1}{x\arcsin x} - \frac{1}{x^2}$$

Using l'hôpital rule, I know the answer is $-\frac{1}{6}$, but it seems like my professor want me to find another way and I can't think of any.

Can you help me?

Answer 1

Hint 1: Convert the given limit expression as a single fraction.

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Hint 2: Use the series expansion for $\sin^{-1}(x)$. i.e., $$\boxed{\sin^{-1}(x) = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \frac{35x^9}{1152} + \dotsc}$$

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Hint 3: Apply the formula, $$\boxed{\lim_{x\to 0}\dfrac{\sin^{-1}(x)}{x} =\lim_{x\to 0}\dfrac{x}{\sin^{-1}(x)} = 1}$$

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Can you take it from here?

Answer 2

Make a substitution $y=\arcsin x \implies x=\sin y$. Thus, the limit becomes (assuming the limit exists):

$$\lim_{y\to0}\frac{\sin y-y}{y\sin^2y}=\lim_{y\to0}\frac{\sin y-y}{y^3} \cdot \lim_{y\to0}\frac{y^2}{\sin^2y}$$

Now let $$L_1=\lim_{y\to0}\frac{\sin y-y}{y^3}$$ Making another substitution $y=3t$, we get (again, assuming the limit exists) $$L_1 = \lim_{t\to0}\frac{\sin(3t)-3t}{27t^3} = \lim_{t\to0} \frac{3}{27} \frac{ \sin(t)-t}{t^3} - \lim_{t\to0} \frac{4}{27} \frac{\sin^3(t)}{t^3} = \frac{1}{9} L_1 - \frac 4{27}$$ $$ L_1=-\frac{1}{6} $$