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anonymous
 one year ago
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anonymous
 one year ago
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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do some of that conjugate business. Remember that stuff?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We need to cancel out the (x1) somehow.. So we need an (x1) to magically appear in the numerator.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \frac{\sqrt{x^2+3}2}{(x1)}\cdot\color{royalblue}{\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)}=\quad?\]Conjugate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would multiply them together?

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{\sqrt{x^2+3}2}{(x1)}}\) \(\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{\sqrt{x^2+3}22x+2x}{(x1)}}\) \(\large \color{blue}{\displaystyle\lim_{x\rightarrow 1}\frac{2x+\sqrt{x^2+3}}{(x1)}\frac{22x}{(x1)}}\) and you can apply L'Hospital's rule to limit #1, and second factors out.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Yes multiply the tops together, leave the bottoms alone though.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \frac{\sqrt{x^2+3}2}{(x1)}\cdot\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)=\frac{\color{orangered}{(\sqrt{x^2+3}2)(\sqrt{x^2+3}+2)}}{(x1)(\sqrt{x^2+3}+2)}\]Leave the bottoms alone like that :) Gotta multiply out this orange stuff though. Remember how to multiply conjugates? Here is our formula: \(\large\rm (ab)(a+b)=a^2b^2\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \left[\sqrt{x^2+3}2\right]\left[\sqrt{x^2+3}+2\right]\quad=\quad\left[\sqrt{x^2+3}\right]^2\left[2\right]^2\]First thing squared, minus, the second thing squared. Which simplifies a bit,\[\large\rm x^2+34\]And a little further, ya? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\rm x^21\) 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \cfrac{x^21}{(x1)(\sqrt{x^2+3}+2)}\implies \cfrac{\cancel{(x1)}(x+1)}{\cancel{(x1)}(\sqrt{x^2+3}+2)}\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Good. So you've turned the equation into:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ive turned it into its conjugate?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \lim\frac{\sqrt{x^2+3}2}{(x1)}\cdot\left(\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}\right)=\frac{(x+1)(x1)}{(x1)(\sqrt{x^2+3}+2)}\]Well you did a whole bunch of work to make an (x1) magically appear in the top! :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Understand why we didn't multiply out the bottom? We were looking for an (x1) in the top so we could cancel stuff out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes I understand now

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And when I plug x=1 into here it still works

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm \lim_{x\to1}\frac{(x+1)}{(\sqrt{x^2+3}+2)}\]You cancelled out the (x1)'s right? No more problems? :O yay!
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