Question

Limit question help

Answer 1

Apparantly The answer is 1 which makes sense

http://openstudy.com/users/dido525#/updates/542390b4e4b0b55e70f229b5

but I have to show the limit exists which is stupid. I don't know how I would do that.

Answer 2

@hartnn , @jim_thompson5910

Answer 3

Clearly if the value of a limit is found the limit must exist, should it not?

Answer 4

Wait wrong question.

Answer 5

@dumbcow

Answer 6

This is the correct question.

Answer 7

oh shoot, haha i was wondering how the heck 1 was answer to first one

Answer 8

Haha xD .

Answer 9

Just let f(x) = x+1

Answer 10

Why?

Answer 11

@geerky42

Answer 12

Question got cut off. It should say "except perhaps at x=0" .

Answer 13

I'm confused about first sentence part, so i could be wrong...

But with f(x) = x+1, one can show that \(\displaystyle \lim_{x\rightarrow0}\left(f(x)+\dfrac{1}{f(x)}\right) \\= \displaystyle\lim_{x\rightarrow0}\left((x+1)+\dfrac{1}{x+1}\right)\\=0+1+\dfrac{1}{0+1}\\=1+1=2\)

f(x) = x+1 satisfies it.
Then you can show that \(\lim_{x\rightarrow0}x+1=\boxed{1}\)

Answer 14

But why x+1? Why not like... x+2?

Answer 15

How could you prove that f(x)=x+1 though? Just because it is one such function that satisfies the properties doesn't mean f must be that function.

Answer 16

well, I am wrong, lesson learned.

Answer 17

im not sure on this one either, as long as f(x) is continuous, then the limit should exist

Answer 18

It doesn't give us that f is continuous either >.<

Answer 19

@geerky42 , you are not wrong but f(x) could be anything as long as f(0) = 1

Answer 20

Basically just need to show that \(\lim_{x\rightarrow1}f(x)=1\) is true for ALL functions that satisfy \(\lim_{x\rightarrow0}\left(f(x)+\dfrac{1}{f(x)}\right)=2\)

That pellet is hard.

Answer 21

No kidding :P .

Answer 22

The only thing I've been able to show so far is that: $$\lim_{x\rightarrow 0}\left(f(x)^k+\frac{1}{f(x)^k}\right)=2$$for any natural number k. Note sure of that helps or not.

Answer 23

Not*

Answer 24

A few people said squaring both sides and taking a root? Not sure how that would help...

Answer 25

is this for pre-calc ?

Answer 26

It's Calc I.

Answer 27

I'm outta here. Not capable enough to help lol...

Answer 28

Thanks for trying :) .

Answer 29

have you been using the epsilon/delta definition for limits in class?

Answer 30

Yes but it's actually not part of the curriculum. Like we wont ever use it in the course. Mind you I'm past Calc I. This is for a friend.

Answer 31

I've done Calc I, II and III and Differential equations. Taking complex analysis atm.

Answer 32

oh ok , i hated that stuff but i thought maybe thats what they wanted you to show here

anyway just prove that value of limit is 1 and it must exist otherwise the other limit which equals 2 would not exist

Answer 33

But that's exactly what I can't show :P . How would I do that? xD .

Answer 34

Cuz apparantly what he did previously isn't right.

Answer 35

http://openstudy.com/users/dido525#/updates/542390b4e4b0b55e70f229b5

Answer 36

hold on i didn't look at that yet

Answer 37

seems correct to me

Answer 38

But apparently my friend's professor said that's not valid because we haven't shown the limit exists.

Answer 39

Its not. He didn't show that the limit exists. So his steps after are invalid. If the limit existed, then he would be able to solve the equation y+ 1/y=2

Answer 40

But how could we even show that? :/ .

Answer 41

The only idea I can think of involves writing f(x) cleverly as a sum or product of things that have limits we know.

Answer 42

People said square both sides. I'm not sure how that helps...

Answer 43

\[\lim_{x \rightarrow 0}(f(x) + \frac{1}{f(x)}) = \lim_{x \rightarrow 0}f(x)+\frac{1}{\lim_{x \rightarrow 0}f(x)} = 2\]
If
lim f(x) does not exist, then statement above is not true

Answer 44

You would only be able to split the lim like that if the limit existed. There are plenty of functions and sequences that dont have limits, but if you add them you get a limit. For example: $$x_n=n,y_n=1-n.$$ Individually, they have no limit as n goes to infinity, but $$x_n+y_n=1,$$ so the limit of x_n+y_n is 1 as n goes to infinity.

Answer 45

We are trying to show if f(x)+1/f(x) does indeed have a limit, that something like what I gave an example of is not possible.