Can Conjugates Simplify Complex Limits as x Approaches Negative Infinity?

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Homework Help Overview

The discussion revolves around evaluating a limit involving square roots as x approaches negative infinity. The original poster is exploring the use of conjugates to simplify the expression but is uncertain about the correct approach given constraints from their teacher.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for handling the limit, including the use of conjugates and dividing by x. Some express confusion over the original poster's approach and seek clarification on the reasoning behind different methods.

Discussion Status

There is an active exploration of different techniques, with some participants suggesting alternatives to using conjugates. Guidance has been offered regarding the importance of absolute values when simplifying square roots, and there is acknowledgment of the need to be cautious with limits that may lead to indeterminate forms.

Contextual Notes

Participants are working under the constraint that derivatives cannot be used, as specified by the teacher. There is also a focus on ensuring that the methods discussed do not lead to indeterminate forms like 0/0.

cabron299
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lim [3 + sqrt(x^2−x+1)]/[sqrt(9x^2−7x) −2x+5]
x−>-infinity

i know the conjugate of the nominator is 3-sqrt(x^2-x+1), but what is the conjugate of the denominator (or do i need to try some other method)?

PS- I cannot solve this with deravatives b/c the teacher tols us not to
 
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I would handle the roots by writing, for example, sqrt(9x^2-7x)=3*|x|*sqrt(1-7x/(9x^2))=3*|x|*sqrt(1-7/(9x)). So the sqrt is 3*|x| times something whose limit is 1. DON'T mess with conjugates. DO keep track of the absolute values.
 
ummm... my teacher said conjugates were the best for limits with a sqrt involved, i don't think we are in the same page (i wasnt able to understand what you did... if u could clarify that would be great =))
thanks in advance
 
Conjugates aren't your best bet here. Too complicated. I'm basically just suggesting you divide numerator and denominator by x and look at the limit of each term. Be careful when you get to limits like sqrt(x^2-x+1)/x. What is the limit of that one?
 
1?( sqrt(x^2-x+1)/(x) simplifying [x*sqrt(1-1/x+1/x^2)]/x, the x´s cancel and I am left off with sqrt(1-1/x+1/x^2)= 1 right?
 
cabron299 said:
1?( sqrt(x^2-x+1)/(x) simplifying [x*sqrt(1-1/x+1/x^2)]/x, the x´s cancel and I am left off with sqrt(1-1/x+1/x^2)= 1 right?

Try it on your calculator. x->negative infinity. The limit is -1, right? The x comes out of the sqrt as |x|. |x|/x=(-1) if x is negative. I told you to be careful of that.
 
oh ok i need to remember the absolute value after x comes out of the sqrt.. so after i have the limits of all the four parts i can sum them up?
thanks a lot!
 
cabron299 said:
oh ok i need to remember the absolute value after x comes out of the sqrt.. so after i have the limits of all the four parts i can sum them up?
thanks a lot!

Yes. After you divide by x you don't get a 0/0 form for the limit. If you did you would need to think about using conjugates to cancel something. But here you don't.
 
thanks a lot! i got the right answer (1/5) is this method trustworthy for most limits?
 
  • #10
If all the terms in your limit have a finite limit and the final result doesn't have an indeterminant form like 0/0, what could go wrong?
 

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