lim x ---infinity of sqrt*(x^2 +x +1)+x

Question

lim x -->-infinity of sqrt*(x^2 +x +1)+x

anonymous · Answer

gimmick is to multiply by the conjugate

anonymous · Answer

i've tried that i can't figure out to do once you get past that

anonymous · Answer

you end up with x+1/sqrt*(x^2 +x+1) -x

anonymous · Answer

$$\frac{x+1}{\sqrt{x^2+x+1}-x}$$

anonymous · Answer

yeah what you wrote

anonymous · Answer

what i don't get is where to go from there

anonymous · Answer

Multiply everything by the conjugate again maybe?

anonymous · Answer

is l'hopital allowed?

anonymous · Answer

you would end up with $$\sqrt{x^2+x+1}-x^2-x / X^2+1$$

anonymous · Answer

which i don't think gets me anywhere

anonymous · Answer

Nvm, that just complicates things.

anonymous · Answer

Is the answer 0?

anonymous · Answer

they only thing i've seen is dividing the numerator and denominator by x but I can't seem to do much once i get there

the answer is -.5 i just need to show the process

anonymous · Answer

the answer is $-\frac{1}{2}$

anonymous · Answer

in the numerator that would give me $$ 1+\frac{ 1 }{ x }$$ and because it is as x --> negative infinity 1/x equals 0 so the numerator is 1

anonymous · Answer

but the denominator is what i can't work out

anonymous · Answer

Yeah I know, so there is another way possibly.

anonymous · Answer

i don't know

anonymous · Answer

you can divide everything by $x$ just don't bring it inside the radical

anonymous · Answer

but how does that help then?

anonymous · Answer

I tried that satellite, I still end up using L'hostipals rules after a while.

anonymous · Answer

$$\frac{x+1}{\sqrt{x^2+x+1}-x}$$
$$=\frac{1+\frac{1}{x}}{-1+\frac{\sqrt{x^2+x+1}}{x}}$$

anonymous · Answer

Yep I got there.

anonymous · Answer

I ended up using L'hospital's rule.

anonymous · Answer

ignore the $\frac{1}{x}$ up top

anonymous · Answer

right because that would approach zero

anonymous · Answer

and 
$$\lim_{x\to -\infty}\frac{\sqrt{x^2+x+1}}{x}=-1$$

anonymous · Answer

Wait why is that -1?

anonymous · Answer

because $x$ is going to $-\infty$ so the denominator is negative, while the numerator is positive

anonymous · Answer

oh and the degress of the terms are equal

anonymous · Answer

Ohh yeah, right.

anonymous · Answer

essentially, yes

anonymous · Answer

Then the rest is simple.

anonymous · Answer

now you get the $-\frac{1}{2}$

anonymous · Answer

yes thanks, and a really stupid question how do i stop this now that we answerd the question

anonymous · Answer

Press close.

anonymous · Answer

ok, thanks