Question

calculate these limits using the ''squeeze theorem''

Answer 1

\[\lim_{n \rightarrow \infty}(1/\sqrt{(n^4+n^2+1)}+1/\sqrt{n^4+n^2+2}+....1/\sqrt{n^4+11n^2)}\]

Answer 2

The squeeze theorem is basically the idea that you can assume that one limit is the same as another because they're close enough.

So if you have 1/x, you can basically say it's the same as 1/(x+1) because as you approach infinity, the 1 doesn't really matter.

Answer 3

thank you, I know the theory about it :D

Answer 4

Haha well I wasn't done thanks :P

Specifically with that, what would be the easiest way to simplify that equation?

Or another way to ask... What is most dominant in that equation as it approaches infinity?

Answer 5

As n->infinity all the terms tend to zero
For first term
\[\lim_{n \rightarrow \infty}\frac{\frac{1}{n^2}}{\sqrt{1+\frac{1}{n^2}+\frac{1}{n^4}}}\]
First term 0
Second term Zero employing squeeze theorem
similarly third term
Hence
\[\lim_{n \rightarrow \infty}f(n)=0\]

Answer 6

An easier way to think about that would be 1/a number that gets infinitely big.

Which approaches zero.

You don't have to actually do any math for this problem, if you understand what's going on.

Answer 7

well, the thing is that I understand this REALLY good, but I wanted to solve it using the squeeze theorem, and you both didn't use it :D

Answer 8

Lol okay fine. Simplify the denominators. :P

Answer 9

You could get away with using 1/n^2 with the squeeze theorem, which coincidentally still goes to zero :P

Answer 10

I'm little bit confused with this 11n^2 :D

Answer 11

I just did 1/sqrt(n^4) which simplifies to 1/n^2 which is bigger than the equation you gave so the limit is also the same. So hah. :P

Answer 12

Oh the 11n^2 doesn't matter in the long run.

Answer 13

I found the first limit to get off a start then
second term is 1/sqrt(first term+1)
we apply squesze theorem to claim this is zero
consequently if we keep on adding terms we can claim them all to be zero in light of squeeze theorem

Answer 14

So basically what he's saying is you apply a limit to each term, and since they're all the same, the limit for the entirety is the same.

And each term is similar, so you can treat them all the same.

Answer 15

thank you notsobright and quantum torch :D

Answer 16

You're welcome. :)