Exercise 3.2

Question

Exercise 2: 
    Calculate $\lim_{n\rightarrow\infty}\bigl(\sqrt{n^2+n}-n\bigr)$.

ghostofgarborg · Accepted Answer

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Note that $ (\sqrt{n^2 + n} - n) = n/(\sqrt{n^2 + n} + n) $, and
    and therefore
    $$
        \frac {1}{2 + \frac{1}{\sqrt n}} = 
        \frac {n}{(n + \sqrt n) + n} <
        \sqrt{n^2 + n} - n <
        \frac{n}{n+n} =
        \frac 1 2 
    $$
    where $(2 + 1/\sqrt n)^{-1} \to \frac 1 2$. By theorem 3.19, $ \lim_{n \to \infty} \sqrt{n^2+n} - n = \frac 1 2$.

Exercise 3.2

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Exercise 3.2

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