# Finding limit at infinity

1. Evaluate lim x$\rightarrow$-$\infty$ $\sqrt{x^2+x+1}$+x.The answer is -$\frac{1}{2}$.

None.

## The Attempt at a Solution

I multiplied by the conjugate first, so it turns into

lim x$\rightarrow$-$\infty$ $\frac{(x^2+x+1)-x^2}{\sqrt{x^2+x+1}-x}$

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

I divide by 1/x on the top, and 1/√x2 on the bottom.

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

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

At this point, this is all the algebra I can do. So now I have to plug in the -$\infty$.

When x goes to -$\infty$ into $\frac{1}{x}$, I get 0. Same with $\frac{1}{x^2}$, I also get 0.

So wouldn't that make my equation

$\frac{1+0}{(√1+0+0)-1}$? My answer would be undefined then, not -$\frac{1}{2}$...

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Dick
Homework Helper
1. Evaluate lim x$\rightarrow$-$\infty$ $\sqrt{x^2+x+1}$+x.The answer is -$\frac{1}{2}$.

None.

## The Attempt at a Solution

I multiplied by the conjugate first, so it turns into

lim x$\rightarrow$-$\infty$ $\frac{(x^2+x+1)-x^2}{\sqrt{x^2+x+1}-x}$

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

I divide by 1/x on the top, and 1/√x2 on the bottom.

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

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

At this point, this is all the algebra I can do. So now I have to plug in the -$\infty$.

When x goes to -$\infty$ into $\frac{1}{x}$, I get 0. Same with $\frac{1}{x^2}$, I also get 0.

So wouldn't that make my equation

$\frac{1+0}{(√1+0+0)-1}$? My answer would be undefined then, not -$\frac{1}{2}$...
x is NEGATIVE. If that's the case then sqrt(f)/x=(-sqrt(f/x^2)). Think about examples with numbers.

1 person
Yes, but how does that apply to my equation?

Even if x is negative, $\frac{1}{x}$ would be so small that it would be insignificant.

Dick
Homework Helper
Yes, but how does that apply to my equation?
You are screwing up a sign with x being negative. You should be getting 1/(-1-1). The sign on the square root is wrong.

Dick
Homework Helper
Even if x is negative, $\frac{1}{x}$ would be so small that it would be insignificant.
No idea what you are talking about. (-1)*sqrt(2) is -sqrt(2*(-1)^2). It's not sqrt(2*(-1)^2).

I think my algebra is right. What step did I do wrong?

Dick
Homework Helper
I think my algebra is right. What step did I do wrong?

Office_Shredder
Staff Emeritus
Gold Member
When you divide by 1/x on the top, and $1/\sqrt{x^2}$ on the bottom you have multiplied your expression by -1.

Dick
When you divide by 1/x on the top, and $1/\sqrt{x^2}$ on the bottom you have multiplied your expression by -1.