Solving Aggravating Limit: x + \sqrt{x^2 + 2x}

[Reedeegi]

Homework Statement

$$\lim_{x\to\,-\infty}x + \sqrt{x^2 + 2x}$$

Homework Equations

The Attempt at a Solution

I tried conjugating by multiplying both the numerator and denominator by $$x + \sqrt{x^2 + 2x}$$, which gave me $$\frac{2x}{x - \sqrt{x^2 + 2x}}$$, but from there I have no idea on what to do.

 

[Dick]

sqrt(x^2+2x)=sqrt((x^2)*(1+2/x))=|x|*sqrt(1+2/x), yes?

 

[Reedeegi]

sqrt(x^2+2x)=sqrt((x^2)*(1+2/x))=|x|*sqrt(1+2/x), yes?

from there, I have limited out and made the algebra equal to $$\frac{2x}{x - \left|x\right|}$$. From there, what should I do?

 

[Dick]

Duh. I missed a sign mistake in your numerator. Can you correct it? Now you have to get rid of the absolute value. How can you do that?

 

[Reedeegi]

Duh. I missed a sign mistake in your numerator. Can you correct it? Now you have to get rid of the absolute value. How can you do that?

Ohhh, I double checked, and it was -2x. And, since its approaching negative infinity, the absolute value becomes --x, or +x. and, with this in mind, the equation becomes $$\frac{-2x}{2x}$$, or -1. And when a limit acts on a constant, the limit is the same as the constant. so, the limit equals -1.

 

[Dick]

The limit is -1 alright, but the function isn't a constant. You already took most of the limit when you replaced sqrt(1+2/x) with 1. And |x|=-x if x<0. So x-|x|=x-(-x) is 2x. Your conclusions are right, but some of your words to describe the reasons are off.