Difference of Roots Limit at Infinity

Question

lim n -> ∞ $(\sqrt{n^2 + 6b} - \sqrt{n^2 - n})$

Applying Limit Laws:

= ((lim n -> ∞ $(\sqrt{n^2}$ )+(lim n -> ∞ $(\sqrt{6b}$ )) - ((lim n -> ∞ $(\sqrt{n^2}$ )-(lim n -> ∞ $(\sqrt{n}$ ))

= (lim n -> ∞ $(\sqrt{6b}$ )) + (lim n -> ∞ $(\sqrt{n}$ ))

= 6b $(\sqrt{n}$ )

= 6b$(\sqrt{∞}$ )

= ∞

However when I check this example on wolfram I get lim n -> ∞ = ${\frac{1}{2}}$

Answer 1

As an alternative, by Binomial series

$$x\to 0 \quad\quad(1+x)^\frac12 = 1 + \tfrac{1}{2}x+o(x)$$

we obtain

$$\sqrt{n^2 + 6b}=\sqrt {n^2}\sqrt{1 + \frac{6b}{n^2}}=n\left(1+\frac{3b}{n^2}+o\left(\frac{1}{n^2}\right)\right)=n+\frac{3b}{n}+o\left(\frac{1}{n}\right)$$

$$\sqrt{n^2 -n}=\sqrt {n^2}\sqrt{1 - \frac{1}{n}}=n\left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)=n-\frac12+o(1)$$

thus

$$\sqrt{n^2 + 6b} - \sqrt{n^2 - n}=n+\frac{3b}{n}+o\left(\frac{1}{n}\right)-n+\frac12+o(1)=\frac12+o(1)\to\frac12$$

Answer 2

$$\lim_{n\to \infty}(\sqrt{n^2+6b}-\sqrt{n^2-n})=\lim_{n\to \infty}\frac{n^2+6b-(n^2-n)}{\sqrt{n^2+6b}+\sqrt{n^2-n}}=\lim_{n\to \infty}\frac{6b+n}{\sqrt{n^2+6b}+\sqrt{n^2-n}}=\\ \lim_{n\to \infty}\frac{\frac{6b}{n}+1}{\sqrt{1+\frac{6b}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{0+1}{\sqrt{1+0}+\sqrt{1-0}}=\frac{1}{2}$$

Answer 3

I'm sorry, but your attempt is a disaster.

First of all, you cannot do $$ \sqrt{n^2+6b}=\sqrt{n^2}+\sqrt{6b} $$ for the simple reason it's false, unless $b=0$. Think to $\sqrt{2}=\sqrt{1+1}$ and $\sqrt{1}+\sqrt{1}=2$. Do you agree that $\sqrt{2}\ne2$?

Next, no law of limit allows you to split that limit, because it's in the so-called “indeterminate form $\infty-\infty$”. You need to transform the sequence so that the limit can be computed; a good strategy is to write $$ \textstyle\sqrt{n^2+6b}-\sqrt{n^2-n}= (\sqrt{n^2+6b}-\sqrt{n^2-n})\dfrac{\sqrt{n^2+6b}+\sqrt{n^2-n}}{\sqrt{n^2+6b}+\sqrt{n^2-n}} $$ (recall $(a-b)(a+b)=a^2-b^2$) that becomes $$ \frac{(n^2+6b)-(n^2-n)}{\sqrt{n^2+6b}+\sqrt{n^2-n}} = \frac{n\left(1+\dfrac{6b}{n}\right)} {n\left(\sqrt{1+\dfrac{6b}{n^2}}+\sqrt{1-\dfrac{1}{n}}\right)} $$ You can now simplify $n$ and apply limit laws.