Question

find the limit

Answer 1

\[\lim_{n \rightarrow \infty} 1 - \frac{ 1 }{ n+2 }\]

Answer 2

@freckles

Answer 3

fun you know 1->1 as n->infty right?

Answer 4

and 1/n->? as n->infty?

Answer 5

1/n = 0

Answer 6

yep yep

Answer 7

you basically have 1-(1/n)

Answer 8

so you have 1-0

Answer 9

oh alright

Answer 10

what about when its 1 - 1/(4n+1)

Answer 11

you still basically have \[1-\frac{1}{4} \cdot \frac{1}{n}\]

Answer 12

4n+1 is basically 4n for really really huge n

Answer 13

so you would have 1-1/4*(0)=1-0

Answer 14

how do you get 1/n

Answer 15

we can also do this the longer way (it isn't too much longer) but it might help you see things better

Answer 16

alright lets do it

Answer 17

\[1-\frac{1}{4n+1} =1-\frac{\frac{1}{n}}{\frac{4n}{n}+\frac{1}{n}}\]
I divided top and bottom by n (on the second fraction)

Answer 18

as n gets big 1/n gets closer to 0
4n/n=4
and again as n gets big 1/n gets closer to 0

Answer 19

\[1-\frac{\frac{1}{n}}{4+\frac{1}{n}} \rightarrow 1-\frac{0}{4+0}=1-\frac{0}{4}=1-0=1\]

Answer 20

pretend we have
\[4+\frac{4n^2+n-1}{9n^2-1}\]

Answer 21

and we want to know the limit as n->inftry

Answer 22

notice the highest exponent on bottom is 2

Answer 23

divide both top and bottom by n^2 (on the second fraction there)

Answer 24

\[4+\frac{\frac{4n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}{\frac{9n^2}{n^2}-\frac{1}{n^2}} =4+\frac{4+\frac{1}{n}-\frac{1}{n^2}}{9-\frac{1}{n^2}}\]
could you figure the limit out as n->infty of this?

Answer 25

we get \[4 + \frac{ 4+0-0 }{ 9-0 }\]

Answer 26

omg you are awesome
so you would 4+(4/9)

Answer 27

like you are doing the same thing to the ones you have above

Answer 28

you look for highest exponent on bottom

Answer 29

and you divide by variable^(what that exponent is)

Answer 30

what about when the bottom exponent is higher than the top exponent and vice versa?

Answer 31

we have f(x)/g(x)

as x->infty

---
if deg(f)>deg(g) the limit does not exist (you can be specific in some cases and say the limit is infinity or negative infty)

if deg(g)>deg(f) the limit is 0

Answer 32

deg(h) means the degree of h

Answer 33

can we do an example of both?

Answer 34

\[\lim_{x \rightarrow \infty}\frac{x^2+1}{x} =\lim_{x \rightarrow \infty}\frac{\frac{x^2}{x}+\frac{1}{x}}{\frac{x}{x}}=\lim_{x \rightarrow \infty}\frac{x+\frac{1}{x}}{1}\]
example of deg(top)>deg(bottom)

Answer 35

can you finish that one now?

Answer 36

which one?

Answer 37

the one I'm doing for example just now

Answer 38

yeah we get \[\frac{ \infty+0 }{ 1 }\]

Answer 39

yeah and infty/1=infty

Answer 40

now vice versa
example of deg(bot)>deg(top)
\[\lim_{x \rightarrow \infty}\frac{x}{x^2+1}\]
here we divide top and bottom by x^2

Answer 41

\[\lim_{x \rightarrow \infty} \frac{\frac{x}{x^2}}{\frac{x^2}{x^2}+\frac{1}{x^2}}=\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{1+\frac{1}{x^2}}\]

Answer 42

you can evaluate this limit?

Answer 43

\[\frac{ 0 }{ 1+0 }\]

Answer 44

which is 0/1 which is 0

Answer 45

so if you have
\[\lim_{n \rightarrow \infty}(1-\frac{2}{3n-1})\]
could you find this limit

Answer 46

\[1-\frac{ \frac{ 2 }{ n }}{ \frac{ 3 }{ n }-\frac{ 1 }{ n } }\]

Answer 47

\[1 - \frac{ 0 }{ 3-0 }\]

Answer 48

1 - 0 = 1

Answer 49

\[1-\frac{ \frac{ 2 }{ n }}{ \frac{ 3n }{ n }-\frac{ 1 }{ n } }\]
type-o above
but yes
you do get
1-(0/(3-0))
1-0
1

Answer 50

anyways I must leave you now

Answer 51

could we do an example of telescoping series

Answer 52

oh man..

Answer 53

I might be able to after 2 hours

Answer 54

I'm sorry :(

Answer 55

will you be online tomorrow

Answer 56

maybe
@rational @ganeshie8 are all good with series and telescoping I believe
if you need help in the meantime

Answer 57

I think even @tanjung has had some recent experience with telescopiong series

Answer 58

and @wio

Answer 59

alright i'll see if they can help thanks

Answer 60

peace

Answer 61

Wait, this isn't a series. It's just a limit, right?

Answer 62

he wanted help with telescoping series also