Question

What is the limit as x approaches negative infinity for the function (sqrt(x^2 + x)) - x ?

Answer 1

infinity

Answer 2

1/2

Answer 3

negative infinity...

Answer 4

nope! its 1/2 as told by him :)

Answer 5

chalo yeh to aata hai

Answer 6

as x goes to MINUS infinity?

Answer 7

yes maam

Answer 8

yes

Answer 9

it doesn't matter if it goes to - or + infinity here :)

Answer 10

really? i believe you but i don't see it. first term goes to infinity. second term also goes to infinity. why 1/2?

Answer 11

maam? oh! u never told ur a prof. sat !

Answer 12

more to it than meets the eye

Answer 13

Thank You

Answer 14

no praveen m born chivalric..i can call neone maam

Answer 15

ah! its cause it not addtion rather subtraction...u need to rationalise and then proceed to pull out an x .

Answer 16

exactly

Answer 17

ah! i'm going to crash , now sat is correct!

Answer 18

whole section in Amit Agarwal's "Differential Calculus" on these types of qstns

Answer 19

pellet! i've not slept for 2 days...i'm flipping out :) okay guys night ! and sat-u stand right :)

Answer 20

1/2 is wrong because as x--> -infinity
\(\sqrt{x^2-x}=-x\sqrt{1-(1/x)}\)

Answer 21

but it is not \[\infty - \infty\] is it \[\infty + \infty\]

Answer 22

yea as i told u sat ur right :)

Answer 23

i still stand by 1/2..how is sat right?

Answer 24

it is true if it goes to +infinity

Answer 25

1/2 is right only wen it tends to + inf. and not -vee inf. just let the value tend to - inf. u'll see it tending to +inf :) salaute sat \./

Answer 26

ema123: answer is finite limit doesn't exist .

Answer 27

ur left with \[1/ [\sqrt{1 + 1/x} + 1]\]

Answer 28

now if u put in -inf into it the 1/x tends to zero doesnt it?

Answer 29

I told you above when you pull the x outside the radical it become -x since your x is negative.

Answer 30

If you multiply times the conjugate then by 1/x / 1/x you get 1/(sqrt(x^2+x)+1) as x-> infit you get 1/2. To find x -> -infit you can instead do as x-> infit of f(-x) which you still get 1/2 because 1/-x still goes to 0

Answer 31

Maybe you believe wolframalpha him?
http://www.wolframalpha.com/input/?i=lim_ {x%5Cto+-%5Cinfty}+%28%5Csqrt{x^2%2Bx}-x%29

Answer 32

http://www.wolframalpha.com/input/?i=lim_ {x%5Cto+-%5Cinfty}+%28%5Csqrt{x^2%2Bx}-x%29

Answer 33

sorry the link is not working :(

Answer 34

see himanshi: u can take these rationalisation only wen there is ambiguity in the value, here its direct saying it tends to inf.

Answer 35

i have a solved exaple in the book

Answer 36

you don't need to continues unless you have something "indeterminate"
if you get infinity by inspection that is the answer.

Answer 37

take a pic and post it.we'll tell if the author is wrong.

Answer 38

but here u get infty - infty so u need to rationalize as its an indeterminate form

Answer 39

thats the mistake ur makin....the form is no more inf-inf, its inf+inf since x->-inf... -(-) is +

Answer 40

no because the limit is as x goes to
\[-\infty\] not
\[\infty\]

Answer 41

get it now

Answer 42

so as x goes to
\[-\infty\] both the first and second term go to
\[\infty\] done

Answer 43

oh yes right..now i c it