$$\lim_{n \to \infty} \frac{\int_{0}^{n-1} \sqrt{x} \,dx}{n^{3/2}}$$
Can someone please provide the step by step solution to this? I somehow can't get a grasp of limits.
Determining $\lim_{n \to \infty} \frac{\int_{0}^{n-1} \sqrt{x}\, dx}{n^{3/2}}$
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limits
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0what do you want? – 2017-02-03
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1I suggest going forward that you use comments to ask questions in regards to solutions posted on your behalf, rather than open a new question. It benefits others who might have the same question to have things in one place and self contained. ;-)) – 2017-02-03
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0Thanks for the suggestion, I'll keep that in mind :) – 2017-02-03
1 Answers
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First note that
$$\int_0^{n-1}\sqrt{x}\,dx=\frac23 (n-1)^{3/2}$$
Then, we have
$$\begin{align} \lim_{n\to \infty}\frac1{n^{3/2}}\int_0^{n-1}\sqrt{x}\,dx&=\frac23\lim_{n\to \infty}\left(\frac{n-1}{n}\right)^{3/2}\\\\ &=\frac23\lim_{n\to \infty}\left(1-\frac1n\right)^{3/2}\\\\ &=\frac23 \end{align}$$