Can Conjugates Simplify Complex Limits as x Approaches Negative Infinity?

[cabron299]

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

 

[Dick]

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.

 

[cabron299]

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

 

[Dick]

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?

 

[cabron299]

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?

 

[Dick]

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.

 

[cabron299]

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 alot!

 

[Dick]

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 alot!

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.

 

[cabron299]

thanks a lot! i got the right answer (1/5) is this method trustworthy for most limits?

 

[Dick]

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?