is equal to :

Question

\(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\)  is equal to:

Answer 1

\(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\) (\(\frac00\) case)

= \(\lim\limits_{x\to 0}\frac{2xsin\sqrt{x^2}-0}{3x^2}\)(By using D.L.H Rule)

=  \(\lim\limits_{x\to 0}\frac{sinx}x=\frac23\) (\(\because\lim\limits_{x\to 0}\frac{sin x}x=1\))

\(\therefore\) \(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\) = 2/3