Question

How do you find the lim n -> infinity ( n/ (n^2-1) ^ (1/2) )?

Answer 1

because the terms in the numerator and the denominator has the same degree, so just take its coefficient of both.
it is 1/1 = 1

Answer 2

Use L'Hopital's rule. Since the limits of the numerator and denominator are both infinity, the limit is an indeterminate form. Then, using L'Hopital's rule,\[\lim_{n \rightarrow \infty} \frac{ n}{ (n^2-1) ^ {1/2} }=\lim_{n \rightarrow \infty} \frac{ \frac{d}{dn}n}{ \frac{d}{dn}(n^2-1) ^ {1/2} }=\lim_{n \rightarrow \infty}\frac{1}{\frac{n}{(n^2-1) ^ {1/2}}}=\lim_{n \rightarrow \infty}\frac{(n^2-1) ^ {1/2}}{n}\]This doesn't seem to be much help, until we realize that the limit is the same as the limit of its reciprocal. The only way that can be true is if the limit is one.

Answer 3

Yes, @RadEn, intuitively it does make sense

@AnimalAin, would it make sense use L'Hopital's rule again if we did not notice that the limit was the same as the limit of its reciprocal? I tried to do that, but then I would be back at the original equation. Kind of cool that it will keep looping, lol

Answer 4

On some problems, you need to apply it more than once. On this problem, as you pointed out, it isn't very productive to iterate.

Answer 5

Thanks @AnimalAin & @RadEn :)

Answer 6

yw

Answer 7

Another method that is a little less exotic: The limit of a product is the product of its limits, so you could square the argument, take the limit of the squared expression (use L'Hopital's rule again), and get a limit of one, which has a square root of one.

Answer 8

Keep going on this stuff. Do math every day.