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Find the limit, $\lim_{x \to 0+} \frac{\int_0^x \sin^{-1} t^2 dt}{\sin(x^2)}$

This came out in my exam and unknowingly, I did the L'Hospital rule and got it correct, but I do not understand how $\int_0^x \sin^{-1} t^2 dt = 0$ ?

The denominator is $0$ as $\sin (0) =0$ and I used LH rule because it is in the form of $\frac{0}{0}$

Using L-H rule on the numerator, it involves the first fundamental theroem of calculus

$\frac{d}{dx} \int_0^x \sin^{-1} t^2 dt = \sin^{-1} x^2$ and I would know how to carry on. So, how do I determine that $\int_0^x \sin^{-1} t^2 dt = 0$ ?

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  • $\begingroup$ Why do you want to "determine that $\int_0^x \sin^{-1} t^2 dt = 0$" (whatever that means)? You solved the exercise without this, didn't you? $\endgroup$ Commented Nov 20, 2022 at 8:43
  • $\begingroup$ The numerator tends to $0$ because $\sin^{-1}$ is bounded. $\endgroup$ Commented Nov 20, 2022 at 8:44
  • $\begingroup$ Or simply by continuity of $x\mapsto\int_0^x\dots$ (which is even differentiable, as you showed). $\endgroup$ Commented Nov 20, 2022 at 8:49

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When you plug in $0$ for the limit you’re plugging in $0$ for $x$ not $t$.

$$\lim_{x \to 0+} \frac{\int_0^x \sin^{-1} t^2 dt}{\sin(x^2)}$$ basically means $$\frac{\int_0^0\sin^{-1} t^2 dt}{\sin(0^2)}=0/0$$

Which is obvious since an integral with the same bounds will always equal $0$. Because it’s $0/0$ then you’d apply le hospital rule and so on the solve the limit.

This means that $\int_0^x \sin^{-1} t^2 dt\neq 0$, but $\int_0^0\sin^{-1} t^2 dt=0$

Hopefully this clears up some misunderstanding.

As a side note the actual integral can be expressed in terms of hypergeometric functions as shown here https://www.wolframalpha.com/input?key=&i=integrate+arcsin%28x%5E2%29+from+0+to+x

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Let $N(x):=\int_0^x \sin^{-1} (t^2)\,\mathrm dt.$

As you showed, $N$ is differentiable. Hence it is continuous, so that $$\lim_{x\to0}N(x)=N(0)=0,$$ which is the only thing you missed to apply L'Hôpital's rule.

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