Complex fraction in numerator help?

[rebeccacats]

1. Evaluate the limit http://www4a.wolframalpha.com/Calculate/MSP/MSP64511d2754i3f4iaefab00001fa62g875680a1ia?MSPStoreType=image/gif&s=44&w=125.&h=45.
2. No formulas
3.The answer is -1/9. I have tried multiplying the top by the conjugate but that seems wrong as there are no square roots involved. I just tried making numerator (not sure what that process is called where you multiply the numerator by the other fractions denominator and so on...) into (3-3+h)/(3+h*3) all over h, but that isn't working.

 

[epenguin]

1. Evaluate the limit http://www4a.wolframalpha.com/Calculate/MSP/MSP64511d2754i3f4iaefab00001fa62g875680a1ia?MSPStoreType=image/gif&s=44&w=125.&h=45.
2. No formulas
3.The answer is -1/9. I have tried multiplying the top by the conjugate but that seems wrong as there are no square roots involved. I just tried making numerator (not sure what that process is called where you multiply the numerator by the other fractions denominator and so on...) into (3-3+h)/(3+h*3) all over h, but that isn't working.

Do what you do in elementary arithmetic when you subtract one fraction from another!

 

[HallsofIvy]

The first thing you should do is the subtraction \frac{1}{3+ h}- \frac{1}{3}

 

[rebeccacats]

1/3+h - 1/3 --> (3-(3+h))/(3*(3+h))-->h/(3(3+h)) :( ?

 

[ciubba]

1/3+h - 1/3 --> (3-(3+h))/(3*(3+h))-->h/(3(3+h)) :( ?

What happened to the negative sign on h? Also, you dropped an h from the denominator

 

[rebeccacats]

I guess it should be distributed after I multiplied by the LCD?

Then I would have -h/(3(3+h)) --> -h/(3(3+h)) *1/h --> -1/(3(3+h)--> -1/(9+0)--> -1/9

 

[epenguin]

You have got the right idea, calculation and result now.
If this is homework you are only nearly there - write this fairly short calculation up according to the pattern of argument you must have been taught or illustrated I.e. this gets by us because 'we know what you mean'.