Question

Could someone help me on this question,

Calculate the limit algebraically:

lim x-> infinity (sqroot(x^2+x) - x)

Answer 1

multiply by
\[\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}\]

Answer 2

clear after that?

Answer 3

Yeah thanks, but what would change if it was asking for limit-> -infinity?

Answer 4

you would recognize that the form would not be
\[\infty-\infty\] but rather
\[\infty +\infty\] and the last for is clearly \[\infty\]

Answer 5

*form

Answer 6

Oh i get it thanks a lot, and once I simplify it would I replace x as 0 to solve?

Answer 7

no you would take the limit as x goes to infinity, not zero. you will end up with a expression with a numerator and denominator, so it should be clear

Answer 8

i can write it if you like, but try it

Answer 9

ah ok, so I got \[x^2 / \sqrt{x^2+x} +x\] and then what would be the next step? lol sorry I'm kinda slow today :P

Answer 10

oh no that is not it!

Answer 11

sorry lets do the algebra

Answer 12

answer should be 1/2 if my answer sheet is right :)

Answer 13

yup - 1/2 is correct

Answer 14

\[(\sqrt{x^2+x}-x)\times \frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}\]

Answer 15

we will get 1/2 eventually
first we get
\[\frac{x}{\sqrt{x^2+x}+x}\]

Answer 16

you had an x^2 in the numerator, but it should be x

Answer 17

ohh ok

Answer 18

check the algebra, you get x^2 +x - x^2 in the numerator, leaves only x

Answer 19

oh that makes sense so far

Answer 20

now we use our eyeballs and think, "numerator is a poly of degree 1, and denominator is not a polynomial, but
\[\sqrt{x^2}\] looks like x so the denominator behaves like
\[2x\] as x goes to infinity
\[\frac{x}{2x}=\frac{1}{2}\] done

Answer 21

if you want to do a bunch of rather silly algebra you can multiply top and bottom by
\[\frac{1}{x}\]being careful to make it look like dividing by x^2 under the radical. it is a pain and not worth writing, but if you have to turn it in it might make your teacher happy

Answer 22

Oh okay, though in the numerator shouldn't it be \[\sqrt{x^2 +x}\] instead of just sqroot(x^2)?

Answer 23

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

Answer 24

\[=\frac{1}{\sqrt{1+\frac{1}{x}}+1}\]

Answer 25

take the limit as x goes to infinity, get
\[\frac{1}{2}\] yes you are right, it is
\[\sqrt{x^2+x}\] in the denominator, that is why i said i was using the eyeball method. ignoring the "x" term

Answer 26

ohhh okay then

Answer 27

in other words thinking
"as x goes to infinity the square root of x squared plus x looks like the square root of x squared
(because the x is unimportant compared to the x squared)
and the square root of x squared looks like x

Answer 28

OH, I get it now, thanks a lot for your help and time, I appreciate it :D

Answer 29

yw