# Homework Help: How to solve this limit w/out a calculator

1. Sep 16, 2009

### 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

2. Sep 16, 2009

### 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.

3. Sep 16, 2009

### cabron299

ummm... my teacher said conjugates were the best for limits with a sqrt involved, i dont think we are in the same page (i wasnt able to understand what you did... if u could clarify that would be great =))

4. Sep 16, 2009

### 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?

5. Sep 16, 2009

### cabron299

1?( sqrt(x^2-x+1)/(x) simplifying [x*sqrt(1-1/x+1/x^2)]/x, the x´s cancel and im left off with sqrt(1-1/x+1/x^2)= 1 right????

6. Sep 16, 2009

### Dick

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.

7. Sep 16, 2009

### 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!

8. Sep 16, 2009

### Dick

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.

9. Sep 16, 2009

### cabron299

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

10. Sep 16, 2009

### 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?