# Thread: Limit help with conjugates (i think) :)

1. ## Limit help with conjugates (i think) :)

There is something im not getting with this limit and its like a roadblock for the rest of the Homework because i been getting similar results on the next few problems.

lim (X-1) / [sr(x+3)-2]
x-1

I assume i must use the conjugate method and mult the top and bottom by (srx+3)+2 but doing that with the dem i get [x-3]-4 on the bottom which cant work because the dem. can't be zero!

since i thought it might be a trick question i tied taking the conj of the top but i end up with the anwser 1/4 which is wrong too.

so how do i do this? thanks

edit really sorry i forgo tthe SR at the end of hte problem

2. ## Re: Limit help with conjugates (i think) :)

Originally Posted by Kevfactor
There is something im not getting with this limit and its like a roadblock for the rest of the Homework because i been getting similar results on the next few problems.

lim (X-1) / [sr(x+3)-2]
x-1

I assume i must use the conjugate method and mult the top and bottom by (srx+3)+2 but doing that with the dem i get [x-3]-4 on the bottom which cant work because the dem. can't be zero!

since i thought it might be a trick question i tied taking the conj of the top but i end up with the anwser 1/4 which is wrong too.

so how do i do this? thanks

edit really sorry i forgo tthe SR at the end of hte problem
$\displaystyle \displaystyle \lim_{x \to 1} \frac{x-1}{\sqrt{x+3}-2}=\displaystyle \lim_{x \to 1} \frac{x-1}{\sqrt{x+3}-2} \cdot \frac {\sqrt{x+3}+2}{\sqrt{x+3}+2}=\displaystyle \lim_{x \to 1} \frac{(x-1)(\sqrt{x+3}+2)}{x+3-4} \Rightarrow$

$\displaystyle \Rightarrow \displaystyle \lim_{x \to 1} (\sqrt{x+3}+2)=4$

3. ## Re: Limit help with conjugates (i think) :)

thanks ya i see what i did now and the other problems

im a bit rusty on factoring( only did it once in precalc) and after a lot of head banging i finally remembered i can drop the brakets if its an addition term. i still didnt remember it on that problem tho.