Question

show that
lim x approaching +infinity ([root ([x sqrt]+x )+x)=1/2

Answer 1

?? can't understand that function

Answer 2

(sqrt root (xsquared +x) +x = 1/2

Answer 3

square root

Answer 4

\[ \sqrt{x^2 + x} + x\] ??

Answer 5

yes

Answer 6

does that not just diverge? unless i misunderstand

Answer 7

i know i am supposed to rationalize

Answer 8

wolf bro says no
http://www.wolframalpha.com/input/?i=lim+x- >inf+sqrt%28x%5E2+%2Bx%29+%2B+

Answer 9

might as well try to multiply by the conjugate

Answer 10

http://www.wolframalpha.com/input/?i=lim+x-%3E+infinity++sqrt%28x%5E2%2Bx%29+%2B+x

Answer 11

yes but after rationalizing dont what to do

Answer 12

There might be some error with the question, the limit clearly diverges to infinity

Answer 13

*I think

Answer 14

- infinity

Answer 15

ok thxs

Answer 16

yeah, this has to diverge...
but...\[\sqrt{x^2+x}+x\cdot{\sqrt{x^2+x}-x\over\sqrt{x^2+x}-x}=
{x^2+x-x^2\over\sqrt{x^2+x}-x}={x\over\sqrt{x^2+x}-x}\]\[{1\over\sqrt{1+\frac1x}-1}\]taking the limit gives\[\frac10\]so we are going to positive infinity it looks like to me...

Answer 17

most likely, your function is
http://www.wolframalpha.com/input/?i=lim+x-%3E+infinity++sqrt%28x%5E2%2Bx%29+-+x

Answer 18

^that makes sense

Answer 19

In that case, Follow Turing Test, instead of - on that left, put + ... everything is almost same.

Answer 20

Thanks everyone .. for making Guru!!

Answer 21

welcome :) well-deserved!

Answer 22

i agree, well earned.

you can solve the actual problem using the same method @TuringTest used on the other function

Answer 23

What is the correct statement?

Answer 24

it doesnt work. it just goes to positive infinity

Answer 25

\[\frac{\sqrt{x^2+x}+x}{1} \cdot \frac{\sqrt{x^2+x}-x}{\sqrt{x^2+x}-x}=\frac{(x^2+x)-x}{\sqrt{x^2+x}-x}\]
\[=\frac{x}{\sqrt{x^2+x}-x} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}}= \frac{\frac{x}{\sqrt{x^2}}}{\frac{\sqrt{x^2+x}}{\sqrt{x^2}}-\frac{x}{\sqrt{x^2}}}\]
|x|=x since x>0
Recall \[\sqrt{x^2}=|x|\]
Ok then so we have
\[\frac{\frac{x}{x}}{\sqrt{1+\frac{1}{x}}-\frac{x}{x}}\]
\[\lim_{x \rightarrow \infty}\frac{1}{\sqrt{1+\frac{1}{x}}-1}=\frac{1}{\sqrt{1+0}-1}=\frac{1}{1-1} \text{ DNE} \]
Sorry had to do it myself lol

Answer 26

that would be positive infinity, not DNE

Answer 27

so infinity is a number ?

Answer 28

infinity does not exist

Answer 29

no, but the limit not existing is different than one that goes to infinity

Answer 30

if it goes to -infty from the left and +infty from the right, then the limit DNE

Answer 31

You can say either one

Answer 32

no you cannot, I assure you

Answer 33

Yes you can
All numbers exist
infinity is not a number so it does not exist

Answer 34

here is the difference:
this one goes to infinity: http://www.wolframalpha.com/input/?i=lim+x-%3E+infinity++sqrt%28x%5E2%2Bx%29+%2B+x

this one does not exist
http://www.wolframalpha.com/input/?i=lim+x-%3E+infinity++sqrt%28x%5E2%2Bx%29+%2B+x

Answer 35

typo, meant this one dne
http://www.wolframalpha.com/input/?i=lim+x-%3E+0++x%2F%7Cx%7C

Answer 36

\[\lim_{x\to\infty}x=+\infty\]\[\lim_{x\to\infty}-x=-\infty\]\[\lim_{x\to0}{x\over|x|}\text{DNE}\]

Answer 37

all different

Answer 38

only if\[\lim_{x\to a^+}f(x)\neq\lim_{x\to a^-}f(x)\]does the limit not exist

Answer 39

give some explain here @TuringTest
http://openstudy.com/study?source=email#/updates/4f8b04ede4b09e61bffc5631

Answer 40

If both sides go to positive infinity you can say the limit is positive infinity but you can also say the limit does not exist because infinity is not a number
Same thing for if both sides go to negative infinity
Sure the positive infinity/negative infinity gives us more information about the behavior of the graph but we can say the limit does not exist because infinity is like totally not a number
And of course we can say the limit does exist when the left limit doesn't equal right limit
But we can not say the limit is infinity if the left limit doesn't equal the right limit.
So in all cases where you have the limit is infinity (+/-) you can say DNE
But not in all cases where you have the limit does not exist can you say the limit is infinity.

Answer 41

infinity is not a number, but if you say that that means\[-\infty=\infty=DNE\]you are saying that all three functions I wrote above have the same limit
try taking that to your professor and see what they tell you
I think you are confusing divergence with non-existence for limits

Answer 42

No that is not what I'm saying

Answer 43

I didn't say -inf=inf

Answer 44

I said if you have -inf then you can say the limit does not exist because -inf is not a number
I said if you have inf then you can say the limit does not exist because inf is not a number
I never ever said -inf=inf

Answer 45

I think we should post this as a separate question and get more opinions
"If the limit of a function is +/-infinity, does that mean that the limit DNE"
lets ask!

Answer 46

this is the definition i know, if we're going by definitions

The limit of f(x) as x approaches a is L

if and only if, given e > 0, there exists d > 0 such that 0 < |x - a| < d implies that
|f(x) - L| < e