# Division of roots

This concerns calculus practice/ limits of sequences and series/ convergence of sequences/ question 4:

Quoting the solution

\begin{align} \lim_{n \to \infty} 2(\sqrt{n+6}\sqrt{n+10}-n) &= 2\lim_{n \to \infty} \frac{(\sqrt{n+6}\sqrt{n+10}-n)(\sqrt{n+6}\sqrt{n+10}+n)}{\sqrt{n+6}\sqrt{n+10}+n} \\ &= 2\lim_{n \to \infty} \frac{(n+6)(n+10)-n^2}{\sqrt{n+6}\sqrt{n+10}+n} \\ &= 2\lim_{n \to \infty} \frac{16n+60}{\sqrt{n+6}\sqrt{n+10}+n} \\ &= 2\lim_{n \to \infty} \frac{16+\frac{60}{n}}{\sqrt{1+\frac{6}{n}}\sqrt{1+\frac{10}{n}}+1} = 16. \end{align}

unquoting

My question is actually only about the last step, where both the nominator and denominator get divided by $\frac{1}{n}$. The guy just pulls $\frac{1}{n}$ into the roots in the denominator. But should it not be, that: $\frac{\sqrt{n+6} \sqrt{n+10}}{n} = \frac{\sqrt{n+6} \sqrt{n+10}}{\sqrt{n^2}} = \sqrt{\frac{n+6}{n^2}}\sqrt{\frac{n+10}{n^2}} = \sqrt{\frac{1}{n} + \frac{6}{n^2}} \sqrt{\frac{1}{n} + \frac{10}{n^2}}$

The difference being that the root expressions converge to 0 instead of 1, and all other parts remaining the same,we have: $2\lim_{n \to \infty} \frac{16+\frac{60}{n}}{\sqrt{ \frac{1}{n}+\frac{6}{n^2}}\sqrt{\frac{1}{n}+\frac{10}{n^2}}+1} = 32$

Can you confirm or am I missing something? Thanks a lot in advance! Note by A Space Bear Smith
3 years, 2 months ago

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This is where you made a mistake:

$\cdots = \frac{\sqrt{n+6} \sqrt{n+10}}{\sqrt{n^2}} = \sqrt{\frac{n+6}{n^2}}\sqrt{\frac{n+10}{n^2}} = \cdots$

Note that $\dfrac{a \times b}{c}$ is not equal to $\dfrac ac \times \dfrac bc$.

- 3 years, 2 months ago

Oh man, thanks a bunch! Such an obvious mistake with arithmetic operations . . . . Have a splendid day!

- 3 years, 2 months ago

Yeah no problem.

When dealing with limits (where $n$ becomes unboundedly large), you can (almost) always try to substitute $n=10^6, 10^7, 10^8,\ldots$ and predict what the limit would be.

- 3 years, 2 months ago