Question

What is the limit of:
root (2x+1) minus root 3
divided by (x-1)

Answer 1

Limit as x approaches `what value` ? :)

Answer 2

as x approaches 1 from the right

Answer 3

\[\Large \lim_{x\to1^+}\frac{\sqrt{2x+1}-\sqrt3}{x-1}\]

Answer 4

We want to start by multiplying the top and bottom by the `conjugate` of the numerator,\[\Large \color{orangered}{\sqrt{2x+1}+\sqrt3}\]k?

Answer 5

yes

Answer 6

Leave the mess alone in the denominator.
We should be able to clean up the top though.

So multiplying conjugates gives us the `difference of squares` right? remember that? :o

Answer 7

Here is how that works:
\[\large (a-b)(a+b) \quad=\quad a^2-b^2\]

But for our problem here, it'll look more like this:\[\large (\sqrt a-\sqrt b)(\sqrt a+\sqrt b) \quad=\quad a-b\]

Answer 8

\[\Large \lim_{x\to1^+}\frac{\sqrt{2x+1}-\sqrt3}{x-1}\color{orangered}{\left(\frac{\sqrt{2x+1}-\sqrt3}{\sqrt{2x+1}-\sqrt3}\right)}\]
Understand how this numerator will multiply out?

Answer 9

Grrr typo... sec

Answer 10

No worries about the typo haha. Will the numerator end up being (2x+1) minus root 9?

Answer 11

yah looks good.

Answer 12

Simplify it down further! :)

Answer 13

simplified to 2(x-1) ?

Answer 14

niceeee

Answer 15

\[\Large \lim_{x\to1^+}\frac{2(x+1)}{(x+1)(stuff)}\]
Looks like you're on the right track :)
See a cancellation from there?

Answer 16

Omg I can't type correctly today lol

Answer 17

\[\Large \lim_{x\to1^+}\frac{2(x-1)}{(x-1)(stuff)}\]

Answer 18

the two (x-1)s will cancel out leaving 2 over all the stuff?

Answer 19

Ya!

And since we done some `cancelling`, we should now check to see if 1 is still giving us a problem.

Answer 20

At the start of the problem, if we tried to plug x=1 directly in, we ended up with the indeterminate form 0/0. <- bad.

What happens if we plug x=1 directly in from where we are now?

Answer 21

I ended up getting 2/ root 5 + root 3
but I don't think that's right

Answer 22

Hmm, yah I think you made a boo boo in there somewhere.\[\Large \frac{2}{\sqrt{2(1)+1}+\sqrt3}\quad = \quad \frac{2}{\sqrt3+\sqrt3}\]

Answer 23

OH I plugged 2 in instead of 1! Okay now I got 2 over root 3 plus root 3. There has to be something further I'm guessing because the book says the answer is root 3 over 3

Answer 24

So adding the terms in the denominator:\[\Large \frac{2}{2\sqrt3} \quad=\quad \frac{1}{\sqrt3}\]Right?

Answer 25

It's usually a good idea to try and get `irrational numbers` out of the denominator when possible.

So, from where we are, we want to move the sqrt3 up to the numerator with multiplication.\[\Large \frac{1}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3} \quad=\quad ?\]

Answer 26

wait, how did you get 2 over 2root3?

Answer 27

2 apples, right? :o when we add them