Division of roots

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

Quoting the solution

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


My question is actually only about the last step, where both the nominator and denominator get divided by 1n \frac{1}{n} . The guy just pulls 1n \frac{1}{n} into the roots in the denominator. But should it not be, that: n+6n+10n=n+6n+10n2=n+6n2n+10n2=1n+6n21n+10n2 \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: 2limn16+60n1n+6n21n+10n2+1=32 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!

link to the question: https://brilliant.org/practice/convergence-of-sequences/?subtopic=sequences-and-limits&chapter=limits&p=4

Note by A Space Bear Smith
3 years, 2 months ago

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This is where you made a mistake:

=n+6n+10n2=n+6n2n+10n2= \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 a×bc \dfrac{a \times b}{c} is not equal to ac×bc \dfrac ac \times \dfrac bc .

Pi Han Goh - 3 years, 2 months ago

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

a space bear Smith - 3 years, 2 months ago

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Yeah no problem.

When dealing with limits (where nn becomes unboundedly large), you can (almost) always try to substitute n=106,107,108,n=10^6, 10^7, 10^8,\ldots and predict what the limit would be.

Pi Han Goh - 3 years, 2 months ago

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