Question

Using the squeeze theorem, find the limit as n approaches infinity:\[x _{n}=\frac{ n-1 }{ n+2 }\] I can get the upper bound as n/n = 1, but what would the lower one be (before I simplify it to 1)?

Answer 1

In order to get a smaller expression, I should decrease the numerator and increase the denominator, right? What would I do here?

Answer 2

maybe this will be useful\[\frac{n-1}{n+2}=\frac{n+2-3}{n+2}=?\]

Answer 3

So just have n-3 on the top?

Answer 4

\[\frac{ ? }{ ? } \le \frac{ n-1 }{ n+2 } \le \frac{ n }{ n }\]

Answer 5

i mean \[\frac{n+2-3}{n+2}=1-\frac{3}{n+2} <1\]
sorry that will not help

Answer 6

\[\frac{n-1}{n+2}\]always is smaller than 1 so we cant find an expression like \[\frac{n-1}{n+2} \ge \frac{a}{b}=1\]

Answer 7

Yes. What would a and b be so that it easily simplifies to 1

Answer 8

well... i mean its impossible to find a , b

Answer 9

We could estimate using inequalities, which I guess is the idea. See the right hand side of my expression with the n/n. I made the numerator larger and the denominator smaller to ensure that this was at least as large as my original equation. I need to do the reverse of this, but everything I try seems up failing.

Answer 10

i'll tell u why u fail to do the reverse...

because when u try to decrease the numerator and increase the denominator final result will be smaller than 1 ... since \(\frac{n-1}{n+2}\) is smaller than 1

Answer 11

so it cant be equal to 1

Answer 12

but what u can do is to write\[\frac{n-1}{n+2} = \frac{n+2-3}{n+2}=1-\frac{3}{n+2}>1-\frac{4}{n+2}\]so\[1-\frac{4}{n+2}<\frac{n-1}{n+2}<1\] is it a acceptable reasoning?

Answer 13

@hartnn
what do u think?

Answer 14

i am new to squeeze theorem,but can we find the lower bound by puting n=-n and again taking limit to infinity?

Answer 15

Thanks @mukushla - that's the sort of thing I was looking for!

Answer 16

yw :)