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anonymous

  • one year ago

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  1. zepdrix
    • one year ago
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    Do some of that conjugate business. Remember that stuff?

  2. zepdrix
    • one year ago
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    We need to cancel out the (x-1) somehow.. So we need an (x-1) to magically appear in the numerator.

  3. anonymous
    • one year ago
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    random medal

  4. zepdrix
    • one year ago
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    \[\large\rm \frac{\sqrt{x^2+3}-2}{(x-1)}\cdot\color{royalblue}{\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)}=\quad?\]Conjugate

  5. anonymous
    • one year ago
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    I would multiply them together?

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

  7. zepdrix
    • one year ago
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    Yes multiply the tops together, leave the bottoms alone though.

  8. zepdrix
    • one year ago
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    \[\large\rm \frac{\sqrt{x^2+3}-2}{(x-1)}\cdot\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)=\frac{\color{orangered}{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}}{(x-1)(\sqrt{x^2+3}+2)}\]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
    • one year ago
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    Okay

  10. anonymous
    • one year ago
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    I can't do it :(

  11. zepdrix
    • one year ago
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    \[\large\rm \left[\sqrt{x^2+3}-2\right]\left[\sqrt{x^2+3}+2\right]\quad=\quad\left[\sqrt{x^2+3}\right]^2-\left[2\right]^2\]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
    • one year ago
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    Yes

  13. zepdrix
    • one year ago
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    \(\large\rm x^2-1\) 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
    • one year ago
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    (x-1)(x+1)

  15. jdoe0001
    • one year ago
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    \(\bf \cfrac{x^2-1}{(x-1)(\sqrt{x^2+3}+2)}\implies \cfrac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)}(\sqrt{x^2+3}+2)}\)

  16. zepdrix
    • one year ago
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    Good. So you've turned the equation into:

  17. zepdrix
    • one year ago
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    ?... really?

  18. anonymous
    • one year ago
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    Ive turned it into its conjugate?

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

  20. zepdrix
    • one year ago
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    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
    • one year ago
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    Yes I understand now

  22. zepdrix
    • one year ago
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    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
    • one year ago
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    And when I plug x=1 into here it still works

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

  25. anonymous
    • one year ago
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    Yes!

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