## anonymous one year ago ...

1. zepdrix

Do some of that conjugate business. Remember that stuff?

2. zepdrix

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

3. anonymous

random medal

4. zepdrix

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

5. anonymous

I would multiply them together?

6. idku

$$\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{\sqrt{x^2+3}-2}{(x-1)}}$$ $$\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{\sqrt{x^2+3}-2-2x+2x}{(x-1)}}$$ $$\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{-2x+\sqrt{x^2+3}}{(x-1)}-\frac{2-2x}{(x-1)}}$$ and you can apply L'Hospital's rule to limit #1, and second factors out.

7. zepdrix

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

8. zepdrix

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

9. anonymous

Okay

10. anonymous

I can't do it :(

11. zepdrix

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

12. anonymous

Yes

13. zepdrix

$$\large\rm x^2-1$$ 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.

14. anonymous

(x-1)(x+1)

15. anonymous

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

16. zepdrix

Good. So you've turned the equation into:

17. zepdrix

?... really?

18. anonymous

Ive turned it into its conjugate?

19. zepdrix

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

20. zepdrix

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.

21. anonymous

Yes I understand now

22. zepdrix

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.

23. anonymous

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

24. zepdrix

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

25. anonymous

Yes!