Find the value of the limit $b_n = n - \sqrt{n^2 + 2n}$ if $(1/n)\rightarrow 0$.

Question

Consider the sequence given by $b_n = n - \sqrt{n^2 + 2n}$. Taking $(1/n)\rightarrow 0$ as given, and using both the Algebraic Limit Theorem, and the fact that if $(x_n)\rightarrow x$ then $(\sqrt{x_n})\rightarrow \sqrt{x}$, show $\lim b_n$ exists and find the value of the limit.

Well, because $(1/n) \rightarrow 0$, the sequence $(n)\rightarrow \infty$. Consider $n=\sqrt{n^2}>\sqrt{n^2 - 2n}$ which implies $n>n-\sqrt{n^2 - 2n}>0 $. So if $(n)\rightarrow \infty$ then so does $(n-\sqrt{n^2 - 2n})\rightarrow \infty$ $\blacksquare$

But after writing I wonder if it is okay for me to assume that $n\in \mathbb{N}$. If $n\notin \mathbb{N}$ then my first statement of $(n)\rightarrow \infty$ is wrong because $n$ could also diverge to $-\infty$.

Answer 1

Hint: Multiply numerator and denominator by $$n+\sqrt{n^2+2n}$$

Answer 2

Set $x=\dfrac 1n\;(x>0)$. You want the limit, as $x\to 0$, of $$\frac1x-\sqrt{\frac 1{x^2}+\frac 2x}=\frac{1-\sqrt{1+2x}}{x}=-\frac{\sqrt{1+2x}-\sqrt 1}{x} $$ This is the opposite of the rate of variation, from $0$, of the function $\sqrt{1+2x}$. Can you end the computation?

Answer 3

Observe that $n - (n^2 + 2n)^{\frac{1}{2}}=n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}$.

By the binomial theorem, it follows that $$ \left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=1+\frac{1}{n}-\frac{1}{2n^2}+o(n^{-2}) $$

So $$ n - n\left( 1 + \frac{2}{n} \right)^{\frac{1}{2}}=n - \left(n + 1 - \frac{1}{2n} + o(n^{-1}) \right)=-1+\frac{1}{2n}+o(n^{-1}) $$

Therefore, $\lim n - (n^2 + 2n)^{\frac{1}{2}} = -1$.