# Thread: Having difficulties with the algebra of this limit, and a quick question about UD lim

1. ## Having difficulties with the algebra of this limit, and a quick question about UD lim

lim x -> 16

(4-√x)/(x-16)

I know you have to do the conjugate (multiply both top and bottom by 4+√x) but then I get (16-x)/(x-16)(4+√x) and i'm not exactly sure how to get the x-16 terms to cancel out, although I know the answer will be negative 1/8. Sorry, i'm not too savvy with algebra.

Furthermore I have a question about limits that don't exist. At what point do you know they don't exist. For example

(2-x)/(x-1)² or (x-1)/x²(x+2) both of which have limits that are undefined/DNE. At what point when attempting to solve them do you realize they don't exist, as opposed to thinking maybe you just haven't done an algebraic trick yet. I know this is pretty ambiguous but if you have any personal tips i'd appreciate it.

And finally, thank you alll sooooooooooooooooo much!!! you're so helpful with all of this and i'm envious of you guys lol. Take care!!!

= -1/8

3. ## Re: Having difficulties with the algebra of this limit, and a quick question about UD

Originally Posted by skinsdomination09
[B]lim x -> 16 (4-√x)/(x-16)
$\frac{4-\sqrt{x}}{x-16}=\frac{-1}{4+\sqrt{x}}$

4. ## Re: Having difficulties with the algebra of this limit, and a quick question about UD

Hint:

$x-16=(\sqrt{x})^2-16=(\sqrt{x}-4)(\sqrt{x}+4)$

Then simplify

Plato beat me to it

nvmd

6. ## Re: Having difficulties with the algebra of this limit, and a quick question about UD

Originally Posted by TheEmptySet
Hint:

$x-16=(\sqrt{x})^2-16=(\sqrt{x}-4)(\sqrt{x}+4)$

Then simplify

Plato beat me to it
UGHH I was thinking so much about the conjugate I neglected to consider the radical trick!! Makes me angry! but thanks!!!!!!

7. ## Re: Having difficulties with the algebra of this limit, and a quick question about UD

Now try:

$\lim_{x\to 16} \, \frac{\sqrt{x}+4}{x-16}$

8. ## Re: Having difficulties with the algebra of this limit, and a quick question about UD

Originally Posted by MaxJasper
Now try:

$\lim_{x\to 16} \, \frac{\sqrt{x}+4}{x-16}$
I definetely think it would be undefined because if I factored the bottom or did the conjugate I would still end up with zero on the bottom although the top term cancels out. 1/0

hopefully i'm not wrong with this one