Proving a limit involving infinity: lim sqrt(x^2+x+1)+x=-1/2

[Jakotheshadows]

I am trying to prove my guess of a limit based on the function's table of values as well as the function's graph.

lim x-> negative infinity (sqrt(x^2 + x + 1) + x) = -1/2

I've tried all the algebra that I can think of for this problem, but I keep getting stuck at incorrect conclusions. If someone could show me how this can be proven I'd appreciate it.

This is the best I can come up with:

multiplying the numerator and denominator of the expression by the conjugate radical:

lim x-> negative infinity ((x + 1)/(sqrt(x^2 + x + 1) - x))
so the numerator is going to negative infinity, and I'm still not sure what to make of the denominator, except that (-x) is going to positive infinity. The radical is what is bothering me, because I can't get rid of it and I think it may be going towards positive infinity as well, but I don't know how I should go about evaluating

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

and I can't really divide by the highest power of x since the square root of x^2 isn't equal to x for x < 0 and the limit is for x going negative. Please help.

 

[stapel]

lim x-> negative infinity (sqrt(x^2 + x + 1) + x) = -1/2

When x gets very large in the negative direction, the square root effectively becomes:

. . . . .\(\displaystyle \sqrt{x^2}\,=\,|x|\,=\,-x\)

Multiplying by the conjugate then gives you:

. . . . .\(\displaystyle \frac{x\, +\, 1}{-x\, -\, x}\)

...which goes to:

. . . . .\(\displaystyle \frac{x}{-2x}\)

Simplify to find the limit value. :wink:

 

[galactus]

You are OK so far with the conjugate thing. But there is a little trick we can use.

After multiplying top and bottom by the conjugate, we get:

\(\displaystyle \lim_{x\to -\infty}\frac{x+1}{\sqrt{x^{2}+x+1}-x}\)

Now, divide by \(\displaystyle \sqrt{x^{2}} \;\ and \;\ -x\) where desirable.

\(\displaystyle \lim_{x\to -\infty}\frac{-1-\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^{2}}}+1}\)

Now, can be seen as \(\displaystyle x\to -\infty\), we get \(\displaystyle \boxed{\frac{-1}{2}}\)

See it?.

 

[Jakotheshadows]

When x gets very large in the negative direction, the square root effectively becomes:

. . . . .\(\displaystyle \sqrt{x^2}\,=\,|x|\,=\,-x\)

Multiplying by the conjugate then gives you:

. . . . .\(\displaystyle \frac{x\, +\, 1}{-x\, -\, -x}\)

is it just me or is part of what you're showing me a zero denominator? particularly the step:
\(\displaystyle \frac{x\, +\, 1}{-x\, -\, -x}\)

 

[Jakotheshadows]

except for the strange looking step from the first post, this is all really helpful.

I didn't know that the square root of x^2 when x is large negative just becomes -x.. It is a little hard for me to wrap my mind around, but hopefully I will get used to it. That knowledge was the necessary missing piece for me to solve the thing, and yes I see the terms going to zero in the second post making the limit -1/2. Thanks guys, I can hopefully move on with my life now and get some rogaine for all the hair I've ripped out of my scalp.

 

[stapel]

is it just me or is part of what you're showing me a zero denominator?

Oops! I accidentally typoed an extra "minus" sign in there (though the next step reflected what I'd meant). I've corrected my earlier post.

Thank you!