What is the limit at infinity for the given expression?

[physics604]

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

Homework Equations

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}<br /> +\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}...

 

[Dick]

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

Homework Equations

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}<br /> +\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.

 

[physics604]

Yes, but how does that apply to my equation?

 

[physics604]

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

 

[Dick]

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]

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).

 

[physics604]

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

 

[Dick]

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

Your algebra is right if x>0. Your algebra is dead wrong if x<0. I've told you.

 

[Office_Shredder]

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.

And you also converted $1/\sqrt{x^2}$ into 1/x when you multiplied the second term. They just plain aren't equal.

 

[physics604]

Okay, thanks I got it.