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\(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\) is equal to: |
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Answer» \(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\) (\(\frac00\) 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 |
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