1st year calc, limits

1. May 12, 2006

m0286

Hello
1 More Question
It says Calculate the following limits:
lim x-> -3 ((1/x+3)+6/(x^2-9)) So what i did was changed x^2-9 into (x-3)(x+3) and then made that the common denominator so I got
1(x-3) + 6/(x-3)(x+3) then the (x-3)'s cancel and your left with 6/(x+3) but then I plug int he -3 and I get 6/0 which is undefined, is that the answer that it doesnt exist or am I doing something wrong? THANKS

2. May 12, 2006

GregA

you're looking for $$lim_{\substack{x\rightarrow -3} \frac{1}{x+3} + \frac{6}{x^2-9}$$...you are correct to look for a common denominator...but you made a mistake at the point where you cancel out terms. What is 1(x-3) + 6 ?

Last edited: May 12, 2006
3. May 12, 2006

m0286

Thanks!

Ohh ok yea I get it so 1(x-3)+6 is x-3+6 so x+3 then the x+3's cancel out and you get 1/(x-3) which is 1/-6... THANKS!!

4. May 12, 2006

cool!