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**1. Evaluate lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\sqrt{x^2+x+1}[/itex]+x.The answer is -[itex]\frac{1}{2}[/itex].**

## Homework Equations

None.

## The Attempt at a Solution

I multiplied by the conjugate first, so it turns into

lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{(x^2+x+1)-x^2}{\sqrt{x^2+x+1}-x}[/itex]

= lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{x+1}{\sqrt{x^2+x+1}-x}[/itex]

I divide by 1/x on the top, and 1/√x

^{2}on the bottom.

lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{\frac{x}{x}+\frac{1}{x}}{\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}+ \frac{1}{x^2} }-\frac{x}{x}}[/itex]

= lim x[itex]\rightarrow[/itex]-[itex]\infty[/itex] [itex]\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}

+\frac{1}{x^2}} -1}[/itex]

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

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

So wouldn't that make my equation

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