Question

This one looks weird:
\[f(x)=\lim_{n \rightarrow \infty} \frac{x^{2n}-1}{x^{2n}+1} \\ \text{ where is } f \text{ continuous ?}\]

Answer 1

i wanna cry ='(

Answer 2

lol I'm thinking about it too...

Answer 3

If this helps or if this is a hint to anyone, this comes after sequences and series discussion in the james stewart 6e calculus book.

Answer 4

i felt series involved hmm

Answer 5

ok can i switch the limit to 0 ?

Answer 6

i guess if we do a substitution

Answer 7

hehe

Answer 8

lol but you wanted to say f(x)=(1-1)/(1+1)=0

Answer 9

not sure if this helps but we can write as
\[f(x)=\lim_{n \rightarrow \infty}\frac{1-\frac{1}{x^{2n}}}{1+\frac{1}{x^{2n}}}\]

Answer 10

\(\lim_{n \to 0}\frac{x^{\frac{1}{2n}}-1}{x^{\frac{1}{2n}}+1} \)

Answer 11

no we cant do what u posted :O

Answer 12

can we ? :O

Answer 13

ok my brain is out of service literally , ill wait until u reveal some solution :)

Answer 14

I wonder if we should just consider some random numbers...
\[x=-1 \\ f(-1)=\lim_{n \rightarrow \infty}\frac{(-1)^n-1}{(-1)^n+1}\]
so this limit wouldn't exist
so it isn't defined at x=-1

Answer 15

but f(-1/2)=-1

Answer 16

interesting ,,,,

Answer 17

i cheated

Answer 18

http://www.wolframalpha.com/input/?i=lim+n-%3Einfty+%28%28x%29%5E%282n%29-1%29%2F%28%28x%29%5E%282n%29%2B1%29

Answer 19

so f(x)=-1 for when -1<x<1

Answer 20

i don't know how to show that though yet
and also i guess wolfram considered all cases when finding limits
and the other limits didn't exist

Answer 21

oh :O

Answer 22

ok then i'll work on this ltr

Answer 23

if n goes to infinity then (x^2n) goes to 0 if -1<x<1

Answer 24

\[\lim \limits_{n\to \infty} x^n = \infty , ~0\]

Answer 25

\[\lim_{n \rightarrow \infty}x^{2n} =0 \\ \text{ for } x \in (-1,1) \\ f(x)=\frac{0-1}{0+1}=-1\]
for x=-1 to x=1

Answer 26

does that mean the function is continuous everywhere except -1 and 1 ?
idk, a real analyst will have a more rigorous argument for proving continuity @eliassaab @ikram002p @zzr0ck3r

Answer 27

\(\frac{x^{2n}-1}{x^{2n}+1} = 1-\frac{2}{x^{2n}+1} \)

Answer 28

im here -.-

Answer 29

\(\Large \frac{x^{2n}-1}{x^{2n}+1} = 1-\frac{2}{x^{2n}+1} \)
ok this should help

Answer 30

and if we consider anything outside..maybe we can try to use l'hospital I guess and see what we come up with
\so x>1...
\[\lim_{n \rightarrow \infty} \frac{x^n \ln(x)-0}{x^n \ln(x)+0}=1 , x>0\]
wait this isn't right is it?

Answer 31

\( \lim 1-\frac{2}{x^{2n}+1} =1-2=-1 \)
done yollaaa

Answer 32

i mean x>1 *

Answer 33

oops forgot the 2

Answer 34

@ganeshie8 no its cont at 1,-1

Answer 35

\[\lim_{n \rightarrow \infty} \frac{2x^{2n} \ln(x)-0}{2x^{2n} \ln(x)+0}=1 , x>1 \]

Answer 36

im here :(

Answer 37

that sounds cool @myininaya i forgot that its a limit

Answer 38

Oh so it is continuous everywhere ?

Answer 39

so I wonder if it is continuous on (-1,1) U (1,inf)

Answer 40

well we can't do that if x<-1

Answer 41

because of the ln part

Answer 42

so we would have to consider something else for x<-1 I guess

Answer 43

x^2n>0 ppl so its cont !

Answer 44

2n even power :O

Answer 45

im i missing something ?

Answer 46

so what its cont :O
whats that cool program xD

Answer 47

so the function was continuous everywhere
i feel like i have been tricked

Answer 48

i stil think it can get discont at x = -1 and 1, not sure hmm

Answer 49

we need to consider n->infinity

Answer 50

its even power xD

Answer 51

If you divide top and bottom by \(x^{-2n}\), It's just a matter of when that is not continuous, I think.

Answer 52

check demo. only

Answer 53

Since \(1^{\infty}\) is an indeterminate, form, I am guessing it has issues at \(x=\pm 1\).

Answer 54

ok i'll tell u something it would be continues and smooth but derivative does not exist , happy now ?

Answer 55

show me analysis argument marki

Answer 56

like this

Answer 57

im too tired

Answer 58

it has contour track :'(

Answer 59

it is just "drawn" continuously without lifting the hand but everyone knows a square wave has jump discontinuity at transitions

Answer 60

with no branch cut

Answer 61

stupid question
how do you find f'(x)?

Answer 62

like can you ignore the limit part for the time being and just use quotient rule?

Answer 63

\[
\lim_{x\to 1^+}f(x) = 1 \neq -1 = \lim_{x\to 1^{-}}f(x)
\]

Answer 64

and then reinsert limit part

Answer 65

wio did u look at the animation ?
how can u say that without fixing n value ?

Answer 66

\[f(x)=\lim_{n \rightarrow a}g(x,n) \\ \text{ \in general what is the rule here for finding derivative } \]

Answer 67

I'm not sure why I want to find the derivative lol

Answer 68

Well \(x\to 1^+\) would mean that we treat limit as if \(x>1\) and \(x^{-2n}\to 0\): \[
\lim_{n\to \infty }\frac{x^{2n}-1}{x^{2n}+1} = \lim_{n\to \infty }\frac{1-x^{-2n}}{1+x^{-2n}} =1
\]

Answer 69

Is that not good enough?

Answer 70

and so since f(1)=0 then it is not continuous at x=1

Answer 71

I don't think you can say \(f(1)=0\).

Answer 72

\(\color{blue}{\text{Originally Posted by}}\) @myininaya
\[f(x)=\lim_{n \rightarrow a}g(x,n) \\ \text{ \in general what is the rule here for finding derivative } \]
\(\color{blue}{\text{End of Quote}}\)
In this case, you differentiate with respect to \(n\).

Answer 73

the sun is rising here finally XD

Answer 74

\[f(1)=\lim_{n \rightarrow \infty}\frac{(1)^{2n}-1}{(1)^{2n}+1}=\lim_{n \rightarrow \infty}\frac{1-1}{1+1}=\frac{0}{2}=0\]
why can't we do this?

Answer 75

Oh, I guess you can do that.

Answer 76

yes yes '!

Answer 77

and you showed \[\lim_{x \rightarrow 1^+}f(x)=1 \neq f(1)\]

Answer 78

so this was a double limit question

Answer 79

my problem is with the smoothly of this function , its so smooth xD

Answer 80

like x=1 function

Answer 81

well in @ganeshie8 picture when the n's were getting larger
the turned at x=1 and x=-1 were getting sharper

Answer 82

the turns*

Answer 83

somehow when u convert smooth dot derivable functions to complex they also continues

Answer 84

so for large n it is not smooth there at x=1 or x=-1

Answer 85

lets try this definition of continuity

no branches cuts ...