# Difference of Roots Limit at Infinity

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 -> ∞ = ${\frac{1}{2}}$

• Do you think that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$? I'm sorry, but nothing in your attempt is salvageable. Jan 14 '18 at 23:54
• @egreg if you write it as $(a + b)^1/2$ then it is $(a^1/2 + b^1/2)$ Jan 14 '18 at 23:55
• $\sqrt{2}=\sqrt{1+1}=\sqrt{1}+\sqrt{1}=2$. Uhm, something's amiss. Jan 14 '18 at 23:56
• You need to go over your high school math textbook and/or notes. You need to fill in the gaps in the basics before starting calculus Jan 15 '18 at 0:00
• @sawreals2 Writing it as $(a+b)^{1/2}=a^{1/2}+b^{1/2}$ does not make it true. It is not true. The square root is not a linear operation. Only linear operations distribute over addition. The only thing that does this is multiplication, differentiation, integration, and a few other well known linear operators. Square roots, powers, logs, trig functions, reciprocals, absolute values DO NOT DO THIS! Jan 15 '18 at 0:04

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

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

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

• What law have you applied here? Jan 15 '18 at 0:07
• $lim_{n \to \infty} \frac{f(n)}{g(n)}= \frac{lim_{n \to \infty} f(n)}{lim_{n \to \infty} g(n)}$ and that for addition of limits and simple algebraic transformations. Jan 15 '18 at 0:12
• How can you apply this if there is no division? Jan 15 '18 at 0:12
• Look at the first equality, on the right hand side there is a division. Jan 15 '18 at 0:18
• I'm sorry but I am failing to see where the division is. Jan 15 '18 at 0:20

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.

• I just don't understand how the quotient rule is applied if there is no division in the original limit I posted. Jan 15 '18 at 0:22
• @sawreals2 The trick is to note that $a=\frac{ab}{b}$, and to choose a suitable $b$, like I did. Jan 15 '18 at 0:23
• How do know when to apply this trick? Wouldn't b just cancel out? Jan 15 '18 at 0:26
• @sawreals2 This is basic stuff (and pretty boring, too). You should have a thorough review of those basics. Jan 15 '18 at 0:33
• Did you factor out n or $n^2$ in the last part? How is 6b/n^2? and 1/n? Jan 15 '18 at 0:47