Question

lim as x approaches negative infinity (x + (x^2 +x+1)^0.5)

Answer 1

\(\large\color{black}{ }\)\[\LARGE \lim_{x \rightarrow \infty }~x+\sqrt{x^2+x+1}\]Right?

Answer 2

maybe it is\[\LARGE \lim_{x \rightarrow \infty }~x\color{red}{-}\sqrt{x^2+x+1}~~~?\]

Answer 3

its plus

Answer 4

to negative infinity

Answer 5

Ohh, my bad

Answer 6

\[\LARGE \lim_{x \rightarrow ~-\infty }~x+\sqrt{x^2+x+1}\]

Answer 7

Sorry about that.

Answer 8

\[\LARGE \lim_{x \rightarrow ~-\infty }~x+\sqrt{x^2+x+1} \]
\[\Large \lim_{x \rightarrow ~-\infty }~(x+\sqrt{x^2+x+1} )\frac{-x+\sqrt{x^2+x+1}}{-x+\sqrt{x^2+x+1}}\]

Answer 9

Rationalizing...

Answer 10

\[\Large \lim_{x \rightarrow ~-\infty }~\frac{x+1}{-x+\sqrt{x^2+x+1}} \]

Answer 11

Then re-write the sum of the 2 limits (hope you know which) separately.

Answer 12

re-write by dividing by x ?

Answer 13

I meant
\[\large \lim_{x \rightarrow ~-\infty }~\frac{x}{-x+\sqrt{x^2+x+1}} +\large \lim_{x \rightarrow ~-\infty }~\frac{1}{-x+\sqrt{x^2+x+1}} \]

Answer 14

you could be done if you divide numerator and denominator by x in this step itself
\(\Large \lim_{x \rightarrow ~-\infty }~\frac{x+1}{-x+\sqrt{x^2+x+1}}\)

Answer 15

yes, likely...

Answer 16

yes, likely...

Answer 17

so 1 / (-1 -inifinity) + zero = zero

Answer 18

no, I don't think this is right,

Answer 19

\(\Large \lim_{x \rightarrow ~-\infty }~\frac{x+1}{-x+\sqrt{x^2+x+1}} \\= \Large\Large \lim_{x \rightarrow ~-\infty }~\frac{1+1/x}{-1+\sqrt{1+1/x+1/x^2}} \)

since x -> - infinity
1/x ->0
1/x^2 ->0

so you can simply plug 0 in place of 1/x and /x^2

Answer 20

ah 1/(1+1) = 1/2

Answer 21

No it is not 1/2 I was about to reply, but as again latex disconnected me.

Answer 22

I lost all of it:)