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Finding limit at infinity

  • Thread starter physics604
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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/√x2 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]...
 

Answers and Replies

  • #2
Dick
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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/√x2 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]...
x is NEGATIVE. If that's the case then sqrt(f)/x=(-sqrt(f/x^2)). Think about examples with numbers.
 
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  • #3
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Yes, but how does that apply to my equation?
 
  • #4
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Even if x is negative, [itex]\frac{1}{x}[/itex] would be so small that it would be insignificant.
 
  • #5
Dick
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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.
 
  • #6
Dick
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Even if x is negative, [itex]\frac{1}{x}[/itex] 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).
 
  • #7
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I think my algebra is right. What step did I do wrong?
 
  • #8
Dick
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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.
 
  • #9
Office_Shredder
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When you divide by 1/x on the top, and [itex] 1/\sqrt{x^2}[/itex] on the bottom you have multiplied your expression by -1.
 
  • #10
Dick
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When you divide by 1/x on the top, and [itex] 1/\sqrt{x^2}[/itex] 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.
 
  • #11
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Okay, thanks I got it.
 

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