Question

...

Answer 1

Do some of that conjugate business.
Remember that stuff?

Answer 2

We need to cancel out the (x-1) somehow..
So we need an (x-1) to magically appear in the numerator.

Answer 3

random medal

Answer 4

\[\large\rm \frac{\sqrt{x^2+3}-2}{(x-1)}\cdot\color{royalblue}{\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)}=\quad?\]Conjugate

Answer 5

I would multiply them together?

Answer 6

\(\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{\sqrt{x^2+3}-2}{(x-1)}}\)

\(\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{\sqrt{x^2+3}-2-2x+2x}{(x-1)}}\)

\(\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{-2x+\sqrt{x^2+3}}{(x-1)}-\frac{2-2x}{(x-1)}}\)

and you can apply L'Hospital's rule to limit #1, and second factors out.

Answer 7

Yes multiply the tops together, leave the bottoms alone though.

Answer 8

\[\large\rm \frac{\sqrt{x^2+3}-2}{(x-1)}\cdot\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)=\frac{\color{orangered}{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}}{(x-1)(\sqrt{x^2+3}+2)}\]Leave the bottoms alone like that :)
Gotta multiply out this orange stuff though.
Remember how to multiply conjugates?
Here is our formula: \(\large\rm (a-b)(a+b)=a^2-b^2\)

Answer 9

Okay

Answer 10

I can't do it :(

Answer 11

\[\large\rm \left[\sqrt{x^2+3}-2\right]\left[\sqrt{x^2+3}+2\right]\quad=\quad\left[\sqrt{x^2+3}\right]^2-\left[2\right]^2\]First thing squared, minus, the second thing squared.
Which simplifies a bit,\[\large\rm x^2+3-4\]And a little further, ya? :o

Answer 12

Yes

Answer 13

\(\large\rm x^2-1\)
And then you have to use your difference of squares formula AGAIN to factor this out. Now in the opposite direction though. It should factor into conjugates.

Answer 14

(x-1)(x+1)

Answer 15

\(\bf \cfrac{x^2-1}{(x-1)(\sqrt{x^2+3}+2)}\implies \cfrac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)}(\sqrt{x^2+3}+2)}\)

Answer 16

Good.
So you've turned the equation into:

Answer 17

?... really?

Answer 18

Ive turned it into its conjugate?

Answer 19

\[\large\rm \lim\frac{\sqrt{x^2+3}-2}{(x-1)}\cdot\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)=\frac{(x+1)(x-1)}{(x-1)(\sqrt{x^2+3}+2)}\]Well you did a whole bunch of work to make an (x-1) magically appear in the top! :)

Answer 20

Understand why we didn't multiply out the bottom?
We were looking for an (x-1) in the top so we could cancel stuff out.

Answer 21

Yes I understand now

Answer 22

I like using this thought process for limits:
Step 1: Plug my limit value directly in. If it causes a problem, oops back up.
Step 2: Do some algebra, cancel stuff out if possible.
Step 3: Repeat Step 1.

You noticed that x=1 gives you a bad denominator,
so you had to do some algebra,
cleaning stuff up, getting rid of that troublesome factor.
then you go right back to step 1, plug x=1 directly in, see if it's a problem still.

Answer 23

And when I plug x=1 into here it still works

Answer 24

\[\large\rm \lim_{x\to1}\frac{(x+1)}{(\sqrt{x^2+3}+2)}\]You cancelled out the (x-1)'s right?
No more problems? :O
yay!

Answer 25

Yes!