Prove that lim $(\sqrt{n^2+n}-n) = \frac{1}{2}$

Solution 1:

From $a^2-b^2=(a+b)(a-b)$ we have

$$\sqrt{n^2+n}-n=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}= \frac{1}{\sqrt{1+\frac{1}{n}}+1}$$

from which the result follows immediately.

Solution 2:

Here is another solution. Remark that $$ \sqrt{n^2+n}-n=\sqrt{n^2+n}-\sqrt{n^2}=\frac{(\sqrt{n^2+n}-\sqrt{n^2})(\sqrt{n^2+n}+\sqrt{n^2})}{\sqrt{n^2+n}+\sqrt{n^2}} $$ Then $$ \sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1}. $$ Since $\lim\limits_{n\to \infty} \sqrt{1+\frac{1}{n}}=1$, for $\delta>0$, exists $N$ ($N>\frac{1}{2\delta+\delta^2}$) such that $1<\sqrt{1+\frac{1}{n}}<1+\delta$, for all $n>N$. Then $$ 2<\sqrt{1+\frac{1}{n}}+1<2+\delta $$ So $$ \frac{1}{2}>\frac{1}{\sqrt{1+\frac{1}{n}}+1}>\frac{1}{2+\delta}. $$ Then $$ \frac{\delta}{4}>\frac{1}{2}-\frac{1}{2+\delta}>\frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1}>0 $$ Let $\varepsilon>0$, define $\delta:=4\varepsilon$. So, for $N>\frac{1}{2\delta+\delta^2}=\frac{1}{8\varepsilon+16\varepsilon^2}$, $$ \varepsilon>\frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1}>0, $$ i.e., $$ \lim\limits_{n\to \infty}\sqrt{n^2+n}-n=\lim\limits_{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{1+1}=\frac{1}{2} $$

Solution 3:

Since this post became active after a long period of time and there are many linked questions, here is a generalization for such kinds of limits. Consider the set of functions in the following form: $$ f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x $$ where $$ n,k \in \Bbb N \\ a_k \in \Bbb R $$

Consider the following expansion: $$ a^n – b^n = (a – b)\left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + \cdots + ab^{n – 2} + b^{n – 1}\right) \tag1 $$

Define a function in the form: $$ g(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} $$

Thus $f(x)$ may be rewritten as:

$$ f(x) = g(x) - x $$

Now using $(1)$ we may rewrite $f(x)$ as: $$ f(x) = \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} \tag2 $$

Now taking the limit of $(2)$ one may obtain: $$ \begin{align*} &\lim_{x\to\infty}\frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n }{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} \\ = &\lim_{x\to\infty}\frac{a_1 + a_2+ \cdots + a_n}{\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1} \\ = &\frac{a_1 + a_2 + \cdots + a_n}{n} \tag3 \end{align*} $$

Or summarizing: $$ \boxed{\lim_{x\to\infty}f(x) = \frac{a_1 + a_2 + \cdots + a_n}{n}} $$

Lets now use that result in your limit: $$ \sqrt{n^2 + n} - n = \sqrt{n(n+1)} - n = \sqrt{(n+0)(n+1)} - n $$ Which gives $a_1 = 0$ and $a_2 = 1$ and the highest power is $2$ meaning $n = 2$ in $(3)$. Thus we get: $$ \boxed{\lim_{n\to\infty}x_n = \frac{a_1 + a_2}{2} = \frac{0 + 1}{2} = {1\over 2}} $$

As desired.