Re: [SPLIT, MOVED] limit, x->infty, sqrt[n^2-1] - sqrt[n^2+n+5]
courteous said:
To nevertheless make use of the thread, I have another one at hand:
\(\displaystyle \lim_{n\to\infty}(\sqrt{n^2-1}-\sqrt{n^2+n+5})\)
For this one, I have no inkling, no slight indication of possible way to solve.
(Though, I am aware of all the general rules of dealing with limits.) Don't take me as a slothful.
Looks like a l'Hospital's rule example. Actually, it doesn't look like one, which is the clue. You need:
something
lim ---------------
n->0 somethingelse
so you just have to make it into a fraction. [Man's gotta do what a man's gotta do. --- Clint Eastwood]
sqrt(n^2-1)-sqrt(n^2+n+5) sqrt(n^2-1)+sqrt(n^2+n+5)
------------------------------- * -------------------------------
1 sqrt(n^2-1)+sqrt(n^2+n+5)
n^2 - 1 - (n^2 + n + 5)
-------------------------
sqrt(n^2-1)+sqrt(n^2+n+5)
n^2 - 1 - n^2 - n - 5
-------------------------
sqrt(n^2-1)+sqrt(n^2+n+5)
5 - n
-------------------------
sqrt(n^2-1)+sqrt(n^2+n+5)
Now divide every term by n (standard trick for n-> inf):
And INSIDE the sqrt(), we divide by n^2:
5/n - 1
----------------------------------
sqrt(1 - 1/n^2) + sqrt(1+1/n+5/n^2)
Now all those fractions -> 0 as n-> inf:
- 1
--------------------
sqrt(1) + sqrt(1)
= -1/2
I was right -- it doesn't look like a l'Hospital's rule case.
